Learning Chinese as a Foreign Language through
Baduk in Korea
Research Proposal
By Sun Dechang
Student ID: 81150093
Dept. of Baduk Studies, Myongji University
2015/6/10
I.Introduction
This Chinese language learning program is designed for the participants from Dept. of Baduk Studies, Myongji University to improve their language learning motivation and self confidence of using the target language.
Constructive learning theories tell that learning by doing is the best policy for a teacher to help his students learn efficiently. The program is going to offer the chances for learners to use the target language through Baduk and around Baduk.
A hypothesis will be described that it can help students overcome the pronunciation and communication problems psychologically. The purpose is to discover and show an efficient design for basic language activities related to both teaching and learning.
II.Background
A.The trends of Chinese learning in Korea and the main problems.
Students in schools are losing their motivation to learn foreign languages and are becoming more demotivated as time goes by. Demotivation has become frequent in schools and the number of demotivated learners is increasing.
De-motivation, as the detrimental forces impeding second language acquisition, has unveiled its diverse appearances to EFL learners worldwide. Shared by both sides were the diminishing role of teachers, the high attribution of confidence deficiency
and the increasing proportion of internal de-motivators. Two de-motivators were unique to Chinese subjects, including learning strategy deficiency and the negative attitude toward target language/culture. Peer pressure was the only de-motivator proper to the Koreans. Consequently, a number of possible inducements were sought to explain the highly culture-specific features of de-motivators, involving ethnic philosophy, native culture/language and educational settings.
B. Constructive learning theories and foreign language learning theories. Since some new research in cognitive psychology to support constructivism, it is widely accepted today by educators, curriculum developers and cognitive psychologists. What is meant by constructivism? The term refers to the idea that learners construct knowledge for themselves---each learner individually (and socially) constructs meaning---as he or she learns. Constructing meaning is learning; there is no other kind. The dramatic consequences of this view are two folds; 1) It is a must to focus on the learner in thinking about learning (not on the subject/lesson to be taught). 2) There is no knowledge independent of the meaning attributed to experiences (constructed) by the learner, or community of learners.
https://www.doczj.com/doc/3712500458.html,parison between language learning and Baduk learning.
A few things stand out when considering the relationship of games and teaching. First of all, teaching is hard. Second, game activities design is hard. Finally, being good at both of these things is incredibly difficult for most educators. But the value of games for the development of pedagogy need not rely on the design and development of games that teach. This is because while teachers are most likely not going to be game designers, they are curriculum and course designers. It is this point of commonality between games and pedagogy that is most fertile for the merging of design strategies. Not only does it offer the opportunity for teachers to merely append their existing skill sets rather than embrace an entire alternate fi eld’s worth of knowledge, but it also allows them to expand upon or re-imagine preexisting materials (courses, programs) rather than begin entirely anew. In this way, incorporating those lessons that games have to teach us about learning and literacy is a bit simpler, as educators are able to deal with a lower barrier of entry (since they are developing new methods from those they are already familiar with) and a shallower learning curve (since they need not learn the extensive intricacies behind the design of a complete game) than they would in designing games for the classroom.
III.Methodology
A.Subject:
The research is about Learning Chinese through Baduk in Korea, and by developing a 16-week credit course for students to choose.
The participants are students and teachers. They are all from Dept. of Baduk Studies, Myongji University. The result will be analysis and compared to some Chinese Confucius Colleges.
B.Procedure
1)Literature reviews and seek the support for the program
2)Show general ideas for the research proposal
3)Better the research planning ideas
4)Prepare the teaching and learning materials
5)Design the lesson plans for the 16-week period curriculum
6)Evaluation of teaching plan and curriculum outline
7)Do some promotion activities for students to enroll
8) A case study will be carried out with a leadership and a teacher.
9)Students will be told the learning methods and the general ideas about what to learn and how to learn.
10)By the end of the semester, all the participants will be tested by two ways: firstly it is about language output by a HSK test. Secondly, it is about language learning interests and confidence by a questionnaire survey.
11)The data will be collected and meanwhile make a comparison to other similar length curriculum and students at the similar level.
12)Do some reflective and write a report for a further research and a better curriculum design
Example for classroom activities:
1)Conversation Practice: Look at the following pictures and describe in Chinese
orally and then ask questions related to them.
About Picture One
a.Who wrote it?
b.What does it mean?
c.Do you agree with them?
About Picture Two
a.Is this a baduk board?
b.What are the names of its nine parts?
c.Do you understand the topic of the passage?
About Picture Three
a.What is the Chinese for Connection?
b.What are the names of the three ways of connections?
c.Would you make some other sentences with these Chinese names?
2)Watching the video and then answer the following questions
Meng Tailin Online Baduk Reviews:
Weiqitv: https://www.doczj.com/doc/3712500458.html,/index/video_play?videoId=55766edb9874f044648b4567
a.Why does he think he did a bad move?
b.What about his emotion during the match?
c.What does he say about his mood?
3)Read the text related to the baduk news and answer the following questions
a.Who are they in the picture?
b.Which is the stronger between Gu Li and Jim Zhixi?
c.Who wins the game?
d.How much money could the winner earn in this tournament?
e.What are they doing?
4)Integrated Tasks( Role Play)
With the help of the Baduk Board, try to use Chinese to introduce it. The key words will be listed on the board.
C.Tools
In seeking evidence of the practice, and the effectiveness of the change in Chinese learning practice, a teacher needs to look at it from different perspectives; we need to employ a triangulation of methods.
This is a simple principle, involving the careful choice of a range of data gathering techniques, each of which might illuminate a different aspect of the same issue: The principle of triangulation:
Also, cross-referencing of data from different methods adds to the overall reliability of the research process.
(See also the section on Triangulation in the component on Qualitative Research by Peter Woods.)
As long as we are aware of the limitations of a particular method, we may use some of the following to help us reflect on our concern:
?observation schedules– of students or themselves;
?audio and video tape recording;
?structured or semi-structured interviews;
?class records;
?statistical indicators;
?field notes;
?an analytic memo;
?sociometry;
?photography;
?repertory grids;
?questionnaires;
D.Hypothesis and analysis
Through baduk, the common topics, which are very familiar with the participants, the Chinese language learning will become more interesting and it will improve the interest and confidence in both pronunciation and communication. The question for teachers, then, is how exactly to take advantage of the benefits Baduk would bring
to the classroom. How do we take the exciting characteristics of this board game, and convert those into pedagogical strategies that will affect real change in the classroom, and make the Chinese learning experience better and more beneficial for students? So, where does that leave the possibilities for baduk integrated in the language classroom, and in what ways can we take advantage of Baduk? Undoubtedly there will be some good activities related to Baduk made for teaching purposes, and as those games prove their worth, they will surely carve out an important niche in teaching and learning.
IV.Expectation
The program will offer a new way for students to learn Chinese as a foreign language and develop a new curriculum. The participants including both teachers and students will benefit from the program. What’s more, it will discover something related to the relationship between the Baduk learning and foreign language learning.
1)Create a syllabus for Learning Chinese through Baduk
Syllabi serve several important purposes, the most basic of which is to communicate the instructor’s course design (e.g., goals, organization, policies, expectations, and requirements) to students. Other functions commonly served by a syllabus include: To convey our enthusiasm for the topic and our expectations for the course
To show how this course fits into a broader context ("the big picture")
To establish a contract with students by publicly stating policies, requirements, and procedures for the course
To set the tone for the course, and convey how we perceive our role as the teacher and their role as students
To help students assess their readiness for the course by identifying prerequisite areas of knowledge
To help students manage their learning by identifying outside resources and/or providing advice
To communicate our course goals and content to colleagues
COMPONENTS DESCRIPTION
Title page Course number and title, semester and year, number
of units, meeting times and location, instructor and TA
information (e.g., name, office, office hours, contact
information)
Course description A brief introduction to the course: scope, purpose and
relevance of the material.
Course objectives Skills and knowledge you want students to gain. Course organization Explanation of the topical organization of the course
Materials Required (and/or optional) books (with authors and
editions), reserve readings, course readers, software,
and supplies with information about where they can be
obtained
Prerequisites and co-requisites Courses students need to have taken before yours (or
at the same time); prerequisite skill sets (e.g.,
programming languages, familiarity with software).
Provide advice on what students should do if they lack
these skills (e.g., drop the course; get outside help;
study supplementary material you will provide)
Course requirements What students will have to do in the course:
assignments, exams, projects, performances,
attendance, participation, etc. Describe the nature and
format of assignments and the expected length of
written work. Provide due dates for assignments and
dates for exams.
Evaluation and grading policy What will the final grade be based on? Provide a
breakdown of components and an explanation of your
grading policies (e.g., weighting of grades, curves,
extra-credit options, the possibility of dropping the
lowest grade)
Course policies and expectations Policies concerning attendance, participation,
tardiness, academic integrity, missing homework,
missed exams, recording classroom activities, food in
class, laptop use, etc. Describe your expectations for
2)
The concept: teaching design is to put the learning elements into a good combination and order according to the curriculum objectives and students level. It is a step to get prepared for teaching plans and teaching ideas. Generally speaking, it includes teaching and learning objectives, teaching focus, main learning tasks, teaching methods, teaching procedure and time arrangement.
3)Learners’ book draft can be finished
4)Getting some experiences and get ready for further research related to teaching Korean as a foreign language through Baduk
V.Schedule and Budget
According to the research procedure, the program will be carried out as the following chart.
Estimated Budget for the Program Learning Chinese through Baduk
iterms price($) amount Total($)
materials
textbooks 10.00 20 200.00 printing 0.50 150 75.00 notebooks 1.00 20 20.00 Teaching materials 2.00 20 40.00 Teachers’ books 15.00 15 225.00
Conference Fee
Curriculum evaluation 200.00 Interviews trips 500.00 Teacher’s pay1000.00 others 50.00
Total 2310.00
Teaching Theories related
The differences between Chinese teaching for native speaker and foreign speaker The features of teaching Chinese for foreign speaker
The competence of communication in Chinese
The learning Goals of Teaching Chinese
The principles of teaching Chinese
The problems we should solve
Language learning and Culture
Teaching aims
Lesson plan
Teaching tools
Presentation
Group learning
Reflective
Tasks design
Classroom activities Pronunciation
Word learning
Grammar
Error and correct
The new method for teaching Chinese Speaking
Evaluation
Test
课 题：复数复习小结 教学目的： 1.理解复数的有关概念；掌握复数的代数表示及向量表示. 2.会运用复数的分类求出相关的复数(实数、纯虚数、虚数等)对应的实参数值. 3.能进行复数的代数形式的加法、减法、乘法、除法等运算. 4.掌握复数代数形式的运算法则及加减法运算的几何意义 教学重点：复数的有关概念、运算法则的梳理和具体的应用． 教学难点：复数的知识结构的梳理 授课类型：新授课 课时安排：1课时 教 具：多媒体、实物投影仪 教学过程： 一、知识要点： 1.虚数单位i :(1)它的平方等于-1，即 2 1i =-; (2)实数可以与它进行四则运算，进行四则运算时，原有加、乘运算律仍然成立 2. i 与－1的关系: i 就是－1的一个平方根，即方程x 2=－1的一个根，方程x 2=－1的另一个根是－i 3. i 的周期性：i 4n+1=i, i 4n+2=-1, i 4n+3=-i, i 4n =1 4.复数的定义：形如(,)a bi a b R +∈的数叫复数，a 叫复数的实部，b 叫复数的虚部全体复数所成的集合叫做复数集，用字母C 表示* 3. 复数的代数形式: 复数通常用字母z 表示，即(,)z a bi a b R =+∈，把复数表示成a +bi 的形式，叫做复数的代数形式 4. 复数与实数、虚数、纯虚数及0的关系：对于复数(,)a bi a b R +∈，当且仅当b =0时，复数a +bi (a 、b ∈R )是实数a ；当b ≠0时，复数z =a +bi 叫做虚数；当a =0且b ≠0时，z =bi 叫做纯虚数；当且仅当a =b =0时，z 就是实数0. 5.复数集与其它数集之间的关系：N Z Q R C . 6. 两个复数相等的定义：如果两个复数的实部和虚部分别相等，那么我们就说这两个复数相等即：如果a ，b ，c ，d ∈R ，那么a +bi =c +di ?a =c ，b =d 一般地，两个复数只能说相等或不相等，而不能比较大小.如果两个复数都是实数，就可以比较大小 只有当两个复数不全是实数时才不能比较大小 7. 复平面、实轴、虚轴： 点Z 的横坐标是a ，纵坐标是b ，复数z =a +bi (a 、b ∈R )可用点Z (a ，b )表示，
英语单词变复数规则 1.1 名词复数的规则变化 1.一般情况加 -s清辅音后读/s/ map-maps 浊辅音和元音后读 /z/ bag-bags /car-cars 2. 以s, sh, ch, x等结尾加 -es 读 /iz/ bus-buses/ watch-watches 3. 以ce, se, ze,等结尾加 -s 读 /iz/ license-licenses 4. 以辅音字母+y结尾变y 为i再加es 读 /z/ baby---babies 1.2 其它名词复数的规则变化 1）以y结尾的专有名词，或元音字母+y 结尾的名词变复数时，直接加s变复数。例如：two Marys the Henrys monkey---monkeys holiday---holidays 2）以o 结尾的名词，变复数时： a. 加s，如：photo---photos piano---pianos radio---radios zoo---zoos； b. 加es，如：potato--potatoes tomato--tomatoes c. 上述a和b两种方法均可，如zero---zeros / zeroes。 3）以f或fe 结尾的名词变复数时： a. 加s，如：belief---beliefs roof---roofs safe---safes gulf---gulfs； b. 去f,fe 加ves，如：half---halves knife---knives leaf---leaves wolf---wolves wife---wives life---lives thief---thieves； c. 上述a和b两种方法均可，如handkerchief: handkerchiefs / handkerchieves。 1.3 名词复数的不规则变化 1）child---children foot---feet tooth---teeth mouse---mice man---men woman---women 注意：由一个词加man 或woman构成的合成词，其复数形式也是-men 和-women，如an Englishman，two Englishmen。但German不是合成词，故复数形式为Germans；Bowman是姓，其复数是the Bowmans。 2）单复同形，如deer，sheep，fish，Chinese，Japanese ，li，jin，yuan，two li，three mu，four jin等。但除人民币的元、角、分外，美元、英镑、法郎等都有复数形式。如：a dollar, two dollars; a meter, two meters。 3）集体名词，以单数形式出现，但实为复数。例如： people police cattle 等本身就是复数，不能说a people，a police，a cattle，但可以说a person，a policeman，a head of cattle, the English，the British，the French，the Chinese，the Japanese，the Swiss 等名词，表示国民总称时，作复数用，如The Chinese are industries and brave.中国人民是勤劳勇敢的。 4）以s结尾，仍为单数的名词，如： a. maths，politics，physics等学科名词，一般是不可数名词，为单数。
网易云课堂_C++程序设计入门(下)_第9单元：白公曾咏牡丹芳，一种鲜妍独“异常”_第9单元 - 作业3：OJ编程 - 使用异常进行复数运算的错误处理... 第9单元?-?作业3：OJ编程?-?使用异常进行复数运算的错误处理 查看帮助 温馨提示： 1.本次作业属于Online Judge题目，提交后由系统即时判分。 2.学生可以在作业截止时间之前不限次数提交答案，系统将取其中的最高分作为最终成绩。 在复数的运算中，练习异常处理 依照学术诚信条款，我保证此作业是本人独立完成的。 通过C++内建的异常类，处理复数除法中除数为0 的问题（5分）题目内容请参见【第9单元 - 作业3说明：【OJ - 使用异常进行错误处理】】 时间限制：500ms内存限制：32000kb #include iostream #include exception #include stdexcept #include limits #include cmath
using namespace std; class MyComplex--2. 创建一个类 MyComplex，用来表示复数。 MyComplex(); MyComplex(double a, double b); friend ostream operator (ostream os, const MyComplex z);--4. 重载流插入运算符，使之可以将复数输出为如下的格式(实部如果是非负数，则不输出符号位；输出时要包含半角左右小括号)：friend istream operator (istream is, MyComplex z);--3. 重载流提取运算符，使之可以读入以下格式的输入(两个数值之间使用空白分隔)，将第一个数值存为复数的实部，将第二个数值存为复数的虚部： MyComplex operator+(const MyComplex secondMyComplex);--加法法则：(a+bi)+(c+di)=(a+c)+(b+d)i； MyComplex operator-(const MyComplex secondMyComplex);--减法法则：(a+bi)－(c+di)=(a－c)+(b－d)i； MyComplex operator*(const MyComplex secondMyComplex);--乘法法则：(a+bi)·(c+di)=(ac－bd)+(bc+ad)i； MyComplex operator-(const MyComplex secondMyComplex);--除法法则：(a+bi)÷(c+di)=[(ac+bd)-(c2+d2)]+[(bc-ad)-(c2+d2)]i. private: double a_;
欢迎阅读 复数基础练习题 一、选择题 1．下列命题中： ①若z＝a＋b i，则仅当a＝0，b≠0时z为纯虚数； ②若(z1－z2)2＋(z2－z3)2＝0，则z1＝z2＝z3； ③x＋y i＝2＋2i?x＝y＝2； ④若实数a与a i对应，则实数集与纯虚数集可建立一一对应关系． 其中正确命题的个数是() A．0B．1 C．2 D．3 2 A 3．a A．2 4． A．a＝1＝－1 5．复数 A 6．设a A．a＝3 2， 7．复数 A 8() A．3＋i 9 A．－3 4＋ 10 A．0 11 A．5－ 12．() A．－10 13．设z1＝3－4i，z2＝－2＋3i，则z1＋z2在复平面内对应的点位于() A．第一象限B．第二象限C．第三象限D．第四象限 14．如果一个复数与它的模的和为5＋3i，那么这个复数是() A.11 5 B.3I C. 11 5＋3i D. 11 5＋23i 15．设f(z)＝z，z1＝3＋4i，z2＝－2－i，则f(z1－z2)＝() A．1－3i B．11i－2 C．i－2 D．5＋5i 16．复数z1＝cosθ＋i，z2＝sinθ－i，则|z1－z2|的最大值为() A．5 B. 5 C．6 D. 6 17．设z∈C，且|z＋1|－|z－i|＝0，则|z＋i|的最小值为()
A ．0 B ．1 C.22 D.12 18．若z ∈C ，且|z ＋2－2i|＝1，则|z －2－2i|的最小值为( ) A ．2 B ．3 C ．4 D ．5 19．(2011年高考福建卷)i 是虚数单位，若集合S ＝{－1,0,1}，则( ) A ．i ∈S B ．i 2∈S C ．i 3∈S D.2i ∈S 20．(2011年高考浙江卷)把复数z 的共轭复数记作z ，i 为虚数单位．若z ＝1＋i ，则(1＋z )·z ＝( ) A ．3－i B ．3＋I C ．1＋3i D ．3 21．化简2＋4i ?1＋i ?2 的结果是( ) A ．222．A 23．A 24．i A ．i 25A ．426A ．i 27．( ) A ．|z 28． 2930点分别是A ，B ，C ，D ，则∠ABC ＋∠ADC ＝________. 31．复数4＋3i 与－2－5i 分别表示向量OA →与OB →，则向量AB →表示的复数是________． 32．已知f (z ＋i)＝3z －2i ，则f (i)＝________. 33．已知复数z 1＝(a 2－2)＋(a －4)i ，z 2＝a －(a 2－2)i(a ∈R )，且z 1－z 2为纯虚数，则a ＝________. 34．(2010年高考上海卷)若复数z ＝1－2i(i 为虚数单位)，则z ·z ＋z ＝________. 35．(2011年高考江苏卷)设复数z 满足i(z ＋1)＝－3＋2i(i 为虚数单位)，则z 的实部是________．
英语名词单复数用法大全 一、规则名词的复数形式：名词的复数形式，一般在单数形式后面加-s 或 -es。现将构成方法与读音规则列表如下： 1、一般情况在词尾加 -s：map-maps, sea-seas, girl-girls, day-days 2、以 s, x, ch, sh 结尾的名词后加 -es：class-classes, box-boxes, watch-watches, dish-dishes 3、以 -f 或 -fe 结尾的词变 -f 和 -fe 为 v再加 -es： l eaf-leaves, thief-thieves, knife-knives, loaf-loaves, wife-wives ；加 -s： belief-beliefs, chief-chiefs, proof-proofs, roof-roofs, gulf-gulfs 4 、以辅音字母加y 结尾的名词，变y 为 i 加 -es ： party-parties, family-families, story-stories, city-cities 5 、以音辅字母加y 结尾的名词，或专有名词以y 结尾的，加-s toy-toys, boy-boys, day-days, ray-rays, Henry-Henrys 6、以辅音字母加 -o 结尾的名词一般加 -es： hero-heroes, Negro-Negroes, potato-potatoes, tomato-tomatoes；不少外来词加 -s： piano-pianos, photo-photos, auto-autos, kilo-kilos, solo-solos 两者皆可： zero-zeros/zeroes, volcano-volcanoes/ volcanos 7 、以元音字母加-o 结尾的名词加-s： radio-radios, bamboo-bamboos, zoo-zoos 8 、以 -th 结尾的名词加-s： truth-truths, mouth-mouths, month-months, path-paths 二、不规则名词复数： 英语里有些名词的复数形式是不规则的，现归纳如下： 1、改变名词中的元音字母或其他形式： man-men, woman-women, foot-feet, goose-geese, mouse-mice 2、单复数相同：sheep, deer, series, means, works, fish, species li, yuan, jin, 3、只有复数形式：ashes, trousers, clothes, thanks, goods, glasses, compasses, contents 4、一些集体名词总是用作复数：people, police, cattle, staff 5、部分集体名词既可以作单数（整体）也可以作复数（成员）： audience, class, family, crowd, couple, group, committee, government, population, crew, team, public, enemy, party 6、复数形式表示特别含义：customs(海关 ), forces( 军队 ), times( 时代 ), spirits( 情绪 ), drinks( 饮料), sands(沙滩 ), papers(文件报纸 ), manners(礼貌 ), looks( 外表 ), brains( 头脑智力 ), greens(青菜), ruins( 废墟 ) 7、表示“某国人”加 -s： Americans, Australians, Germans, Greeks, Swedes, Europeans；单复 数同形 Swiss, Portuguese, Chinese, Japanese；以 -man 或-woman 结尾的改为 men,-women，Englishmen, Frenchwomen 8、合成名词将主体名词变为复数：sons-in-law, lookers-on, passers-by, story-tellers, boy friends ； grown-ups, housewives, stopwatches ；将两部分变为复无主体名词时将最后一部分变为复数： 数： women singers, men servants
复数基础练习题 一、选择题 1．下列命题中： ①若z ＝a ＋b i ，则仅当a ＝0，b ≠0时z 为纯虚数； ②若(z 1－z 2)2＋(z 2－z 3)2＝0，则z 1＝z 2＝z 3； ③x ＋y i ＝2＋2i ?x ＝y ＝2； ④若实数a 与a i 对应，则实数集与纯虚数集可建立一一对应关系． 其中正确命题的个数是( ) A ．0 B ．1 C ．2 D ．3 2．在复平面内，复数z ＝sin 2＋icos 2对应的点位于( ) A ．第一象限 B ．第二象限 C ．第三象限 D ．第四象限 3．a 为正实数，i 为虚数单位，z ＝1－a i ，若|z |＝2，则a ＝( ) A ．2 B. 3 C. 2 D ．1 4．(2011年高考湖南卷改编)若a ，b ∈R ，i 为虚数单位，且a i ＋i 2＝b ＋i ，则( ) A ．a ＝1，b ＝1 B ．a ＝－1，b ＝1 C ．a ＝－1，b ＝－1 D ．a ＝1，b ＝－1 5．复数z ＝3＋i 2对应点在复平面( ) A ．第一象限内 B ．实轴上 C ．虚轴上 D ．第四象限内 6．设a ，b 为实数，若复数1＋2i ＝(a －b )＋(a ＋b )i ，则( ) A ．a ＝32，b ＝12 B ．a ＝3，b ＝1 C ．a ＝12，b ＝32 D ．a ＝1，b ＝3 7．复数z ＝12＋12i 在复平面上对应的点位于( ) A ．第一象限 B ．第二象限 C ．第三象限 D ．第四象限 8．已知关于x 的方程x 2＋(m ＋2i)x ＋2＋2i ＝0(m ∈R )有实根n ，且z ＝m ＋n i ，则复数z 等于( ) A ．3＋i B ．3－I C ．－3－i D ．－3＋i 9．设复数z 满足关系式z ＋|z |＝2＋i ，那么z 等于( ) A ．－34＋i B.34－I C ．－34－i D.34＋i 10．已知复数z 满足z ＋i －3＝3－i ，则z ＝( ) A ．0 B ．2i C ．6 D ．6－2i 11．计算(－i ＋3)－(－2＋5i)的结果为( ) A ．5－6i B ．3－5i C ．－5＋6i D ．－3＋5i 12．向量OZ 1→对应的复数是5－4i ，向量OZ 2→对应的复数是－5＋4i ，则OZ 1→＋OZ 2→对应的复数是( ) A ．－10＋8i B ．10－8i C ．0 D ．10＋8i 13．设z 1＝3－4i ，z 2＝－2＋3i ，则z 1＋z 2在复平面内对应的点位于( ) A ．第一象限 B ．第二象限 C ．第三象限 D ．第四象限 14．如果一个复数与它的模的和为5＋3i ，那么这个复数是( ) A.11 5 B.3I C.11 5＋3i D.11 5＋23i 15．设f (z )＝z ，z 1＝3＋4i ，z 2＝－2－i ，则f (z 1－z 2)＝( ) A ．1－3i B ．11i －2 C ．i －2 D ．5＋5i 16．复数z 1＝cos θ＋i ，z 2＝sin θ－i ，则|z 1－z 2|的最大值为( ) A ．5 B. 5 C ．6 D. 6 17．设z ∈C ，且|z ＋1|－|z －i|＝0，则|z ＋i|的最小值为( ) A ．0 B ．1 C.22 D.1 2 18．若z ∈C ，且|z ＋2－2i|＝1，则|z －2－2i|的最小值为( ) A ．2 B ．3 C ．4 D ．5 19．(2011年高考福建卷)i 是虚数单位，若集合S ＝{－1,0,1}，则( ) A ．i ∈S B ．i 2∈S C ．i 3∈S D.2 i ∈S 20．(2011年高考浙江卷)把复数z 的共轭复数记作z ，i 为虚数单位．若z ＝1＋i ，则(1＋z )·z ＝( ) A ．3－i B ．3＋I C ．1＋3i D ．3
一般情况加-s 1.清辅音后读/s/; map-maps 2.浊辅音和元音后bag-bags 读/z/; car-cars 以s,sh,ch, x等结尾的词加-es 读/iz/ bus-buses ，watch-watches 以ce,se,ze, (d)ge等结尾的词加-s 读/iz/ license-licenses 以辅音字母+y 变y 为i 结尾的词再加es 读/z/ baby---babies 名词复数的不规则变化 1）child---children foot---feet tooth---teeth mouse---mice man---men woman---women 注意：与man 和woman构成的合成词，其复数形式也是-men 和-women。 如：an Englishman，two Englishmen. 但German不是合成词，故复数形式为Germans；Bowman是姓，其复数是the Bowmans。 2）单复同形如： deer，sheep，fish，Chinese，Japanese li，jin，yuan，two li，three mu，four jin 但除人民币元、角、分外，美元、英镑、法郎等都有复数形式。如: a dollar, two dollars; a meter, two meters 3）集体名词，以单数形式出现，但实为复数。 如：people police cattle 等本身就是复数，不能说a people，a police，a cattle，但可以说a person，a policeman，a head of cattle,the English，the British，the French，the Chinese，the Japanese，the Swiss 等名词，表示国民总称时，作复数用。如：The Chinese are industries and brave. 中国人民是勤劳勇敢的。 4）以s结尾，仍为单数的名词，如：a. maths，politics，physics等学科名词，为不可数名词，是单数。 b. news 是不可数名词。 c. the United States，the United Nations 应视为单数。 The United Nations was organized in 1945. 联合国是1945年组建起来的。 d. 以复数形式出现的书名，剧名，报纸，杂志名，也可视为单数。 "The Arabian Nights" is a very interesting story-book. <<一千零一夜>>是一本非常有趣的故事书。 5) 表示由两部分构成的东西，如：glasses (眼镜) trousers, clothes 若表达具体数目，要借助数量词pair(对，双); suit(套); a pair of glasses; two pairs of trousers 6）另外还有一些名词，其复数形式有时可表示特别意思，如：goods货物，waters 水域，fishes（各种）鱼
《复数的四则运算》教学设计 吕叔湘中学 黄国才 【教学目的】1、初步理解复数的加法、减法、乘法的运算法则. 2、会利用加法、减法、乘法、运算法则进行简单的运算。 3、了解复数中共轭复数的概念 【教学重点】：会利用加法、减法、乘法、运算法则进行简单的运算。 【教学难点】：理解复数的加法、减法、乘法的运算法则. 【教学过程】： 一、 问题情景： 问题1： 由初中学习我们可以知道： （2+3x ）+(1-4x)=3-x 猜想： （2+3i ）+(1-4i)= ？ 二、 建构数学 1、复数减法的运算法则 问题 2：用字母表示数，你可以表示复数的运算法则和运算律吗？ (1)运算法则:设复数z 1=a+bi,z 2=c+di,（a,b,c,d ∈R ）那么： z 1+z 2=(a+bi)+(c+di)=(a+c)+(b+d)i; 显然，两个复数的和仍是一个复数，复数的加法法则类似于多项式的合并同类项法则。 (2)复数的加法满足交换律、结合律,即对任何z 1,z 2,z 3∈C,有： z 1+z 2=z 2+z 1, (z 1+z 2)+z 3=z 1+(z 2+z 3) 2、复数减法的运算法则 定义：把满足(c+di )+(x+yi) = a+bi 的复数x+yi （x,y ∈R ），叫做复数a+bi 减去复数c+di 的差，记作：x+yi ＝(a+bi )－(c+di) 由复数的加法法则和复数相等定义，有c+x=a ， d+y=b 由此，x=a －c ， y=b －d ∴ (a+bi )－(c+di) = (a －c) + (b －d)i 显然，两个复数的差仍然是一个复数 由此可见： 两个复数相加(减)就是把实部与实部，
北师大版数学选修2-2第五章数系的扩充与复数的引入 自我总结卷 一、选择题: 1、复数1z i =+(i 是虚数单位),则复数(1)(1)z z +-虚部是( ）【答案】D A 、－1+2i B 、－１?Ｃ、2i Ｄ、2 １、0=a 是复数(,)a bi a b R +∈为纯虚数的( ) 【答案】 Ｂ A 、充分条件 B 、必要条件 C 、充要条件 D 、非充分非必要条件 1、已知复数134z i =+，2z t i =+，且12z z 是实数，则实数t 等于( A )．期中考试题 A.错误! B.错误! C.-错误! Ｄ.－3 4 解析 z 1·错误!=(3＋４i)（t -i)＝（3ｔ＋4)＋(4t －3)i ．因为ｚ1·错误!是实数,所以4t -3=0，所以t ＝错误!.因此选A ． 1、若复数22(34)(56)m m m m i --+--是虚数,则实数m 满足( ）【答案】D （A ）1m ≠- (B ）6m ≠ （C) 1m ≠-或6m ≠ （D ） 1m ≠-且6m ≠ 1、若12,z z C ∈,则1212z z z z ?+?是( ） 【答案】B A 纯虚数 B 实数 C 虚数 D 无法确定 1、若22(1)(32)x x x i -+++是纯虚数，则实数x 的值是( ) 【答案】Ａ A 1 B 1- Ｃ 1± Ｄ 以上都不对 1．已知复数1 122 2i,34i,z z m z z =+=-若 为实数，则实数m 的值为( ) 【答案】D A 、２ Ｂ．2- C 、 2 3 ?D.23- 2．i 表示虚数单位，则2008321i i i i ++++ 的值是（ ) 答案 Ａ A.0 B ．1 C.i D.i - 2、已知 z =则501001z z ++的值为(A )
名词复数变化规则和读音： 1．一般在名词的词尾加“s”，清辅音后读/s/ 浊辅音和元音后读/z/ 例如：books, pens, classrooms，map-maps，boy-boys，girl-girls，pen-pens等等。cats 猫rooms 房间horses 马trees 树roses 玫瑰 2．以s, x, sh, ch结尾的名词在词尾加“es”，读/iz/ 例如：classes, boxes, brushes, dishes, watches, buses，classes，foxes ，lashes 鞭子，pushes 推力，branches 树枝、分支，matches 火柴、比赛，coaches 教练，gases 气体，asses 驴子但也有例外，如：stomach—stomachs等等。 3．以辅音字母加“y”结尾的名词应改“y”为“i”，再加“es”，读/z/ 例如：cities, universities)，factories，baby---babies，city-cities，country-countries，families 家庭，ponies 小马 但以y结尾的专有名词，或元音字母+y 结尾的名词变复数时，直接加s变复数。 例如：two Marys，the Henrys，monkey---monkeys，holiday---holidays，boys, toys 4．以“f”和“fe”结尾的名词应改“f”和“fe”为“ves” 例如：shelf—shelves架子，knife—knives，leaf---leaves叶，wolf---wolves狼，wife---wives 妻子life---lives，thief---thieves小偷，calf---calves 小牛，half---halves 一半但也有例外 如：roof—roofs屋顶，cliff—cliffs（悬崖），hoof—hoofs（马蹄），belief—beliefs（信仰），chief—chiefs（首领），proof—proofs(证明)，safe—safes(保险箱)，reef—reefs(礁)，gulf---gulfs 海湾 还有一些该类名词的复数形式有两种变化形式的 例如：scarf—scarfs/scarves（头巾）, dwarf—dwarfs/dwarves（矮子）, wharf—wharfs/wharves(码头), handkerchief—handkerchiefs/handkerchieves(手帕)等等。 5．以“o”结尾的名词的复数形式一般在词尾加“es” 例如：hero—heroes英语, echo—echoes(回音)，potato—potatoes马铃薯，tomato—tomatoes西红柿，mango---mangoes 芒果，volcano---volcanoes 火山，negro---negroes 黑人， cargo---cargoes 货物，buffalo---buffaloes 水牛，mosquito---mosquitoes 蚊子 但以字母o结尾的外来词或缩写词的复数形式是只加S， 例如：zoo—zoos, radio—radios, piano—pianos钢琴, photo—photos, memo—memos(备忘录), solo—solos(独唱、独奏), kilo—kilos(公斤)，kimono—kimonos（和服），bamboo---bamboos 竹子，kangaroo---kangaroos 袋鼠，mulatto---mulattos 白黑混血儿， 6. 如果名词结尾是一个元音(即a,e,i,o,u) 加y，那只则在单数词后加一个s就行了。 play plays 戏剧 way ways 小路 valley valleys 山谷 donkey donkeys 驴
复数基础——复数的基本运算_2 回顾复数 复数的基本运算 回顾复数 将下列数字写成复数形式： ?简单复习一下，复数是包含实数部分和虚数部分的数。 如果有a+bi，a是实数，b是实数，这是复数。a是实部，bi是虚数部分（注：虚部不包括i）。 为什么bi是虚部？因为bi带有特殊系数i，这个虚数单位，这个特殊的数i，在这里乘以了b。我相信大家都会觉得怪诞，不过根据定义：?在此之前，不存在对某个数取平方后得到-1，现在取i的平方，得到-1，关于虚数（单位）的特别的知识点是它的平方是负数。复数有用之处在于它使我们有能力解决很多方程，这些方程在只允许实数解的情况下无解。复数在很多方面都有用，特别是在工程领域，还有其他领域，比如物理等等。现在，我们不会花很多心思讨论复数定义，在大家处理更多数字后，特别是接触到某些工程应用后，希望大家明白虚数的价值。 回到问题中来，把上面的数字写成复数形式。 ?怎么把它写成复数呢？把它写成实部和虚部的组合。可以写成： -21 = -21+0i ?0i等于0，所以它仍等于-21，实际上这里没有虚部，-21本身就是复数形式，很简单。同样的：
7i是虚数形式的，所以这里没有实部，实部是0，虚部是7i，所以等于0 + 7i。 复数的基本运算 很多时候解方程都会碰到根号下负数的情况，比如根号下-1或者-9：由于如何实数的平方不是0就是正数，所以以上两个数这些没有定义，为了定义这些数，人们引入i的概念，i是虚数单位，i的定义是：这就是解决了根号下负数的问题，这样一来，根号下-9是多少呢？它等于i乘以根号9，即3i， 为什么，想想3i平方是多少？ 这是指数性质。所以，这样的定义就拓展到了，所有负数开根号的情况： 3i是所谓的虚数，它其实也不比其他数“虚”，某种意义上，负数真的存在吗？只不过是将负号放在前面表示抽象含义，负号只是表示它和大小的关系。任何数乘以虚数单位i都是虚数。解二次方程时，你会发现结果有时会实数和虚数并存（有实数部分和虚数部分），举个例子：这不能化简了，因为实数和虚数不能相加，大家可以把这当作不同维度，一个数有实部5，还有虚部2i，这叫做复数。复数可以在平面中表示：虚数也就是虚轴，在纵轴2i，上图表示为2个单位。 实数也就是实轴，在实轴5，上图表示为5个单位。 所以这个图形表示为：5+2i。在以后讲复数应用时，我还会举更多例子，现在只需要知道定义即可。看看有什么运算，两复数相加怎么做：a是实部，bi是虚部，另一个复数是：
复 数 复数的基本概念、复数相等的充要条件以及复数的代数运算是高考的热点，并且一般在前三题的位置，主要考查对复数概念的理解以及复数的加减乘除四则运算，难度较小． 【复习指导】 1．复习时要理解复数的相关概念如实部、虚部、纯虚数、共轭复数等，以及复数的几何意义． 2．要把复数的基本运算作为复习的重点，尤其是复数的四则运算与共轭复数的性质等．因考题较容易，所以重在练基础。 基础梳理 1．复数的有关概念 (1)复数的概念 形如a ＋b i (a ，b ∈R )的数叫作复数，其中a ，b 分别是它的实部和虚部．若b ＝0，则a ＋b i 为实数，若b ≠0，则a ＋b i 为虚数，若a ＝0且b ≠0，则a ＋b i 为纯虚数． (2)复数相等：a ＋b i ＝c ＋d i ?a ＝c 且b ＝d (a ，b ，c ，d ∈R )． (3)共轭复数：a ＋b i 与c ＋d i 共轭?a ＝c ，b ＝－d (a ，b ，c ，d ∈R )． (4)复平面 建立直角坐标系来表示复数的平面，叫作复平面．x 轴叫作实轴，y 轴叫作虚轴．实轴上的点都表示实数；除原点外，虚轴上的点都表示纯虚数；各象限内的点都表示非纯虚数． (5)复数的模 向量OZ →的模r 叫作复数z ＝a ＋b i 的模，记作__|z |__或|a ＋b i|，即|z |＝|a ＋b i|＝a 2＋b 2. 2．复数的几何意义 (1)复数z ＝a ＋b i(a ，b ∈R )的模|z |＝a 2＋b 2，实际上就是指复平面上的点Z 到原点O 的距离；|z 1－z 2|的几何意义是复平面上的点Z 1、Z 2两点间的距离． (2)复数z 、复平面上的点Z 及向量OZ → 相互联系，即z ＝a ＋b i(a ，b ∈R )?Z (a ，b )?OZ → . 3．复数的四则运算 设z 1＝a ＋b i ，z 2＝c ＋d i(a ，b ，c ，d ∈R )，则 (1)加法：z 1＋z 2＝(a ＋b i)＋(c ＋d i)＝(a ＋c )＋(b ＋d )i ； (2)减法：z 1－z 2＝(a ＋b i)－(c ＋d i)＝(a －c )＋(b －d )i ； (3)乘法：z 1·z 2＝(a ＋b i)·(c ＋d i)＝(ac －bd )＋(ad ＋bc )i ； (4)除法：z 1z 2 ＝a ＋b i c ＋d i ＝(a ＋b i )(c －d i )(c ＋d i )(c －d i )＝(ac ＋bd )＋(bc －ad )i c 2＋d 2(c ＋d i ≠0)．
复数的三角形式的运算(一)·教案示例 目的要求 1．掌握复数三角形式的乘法运算法则． 2．理解复数三角形式的乘法运算的几何意义，并能简单地应用． 内容分析 1．在代数形式下，两个复数的乘积(a ＋bi)(c ＋di)按照多项式展开，从而得出乘法运算法则．在三角形式下，两个复数的乘积r1(cos θ1＋isin θ1)·r2(cos θ2＋isin θ2)仍可按代数形式(r1cos θ1＋ir1sin θ1)(r2cos θ2＋ir2sin θ2)来计算．但这样运算较繁杂，而且没有体现出三角形式下模与辐角的特征和作用，因此很有必要研究两个复数的乘积的结果(也是一个复数)的模与原来两个复数的模、辐角与原来两个复数的辐角之间的关系． 2．三角形式下两个复数z1＝r1(cos θ1＋isin θ1)与z2＝r2(cos θ2＋isin θ2)的乘法公式及法则： r1(cos θ1＋isin θ1)·r2(cos θ2＋isin θ2)＝r1r2[cos(θ1＋θ2)＋isin(θ1＋θ2)] 即，两个复数相乘，积的模等于各复数的模的积，积的辐角等于这两个复数的辐角的和． 上述法则中，注意“积的辐角等于这两个复数的辐角的和”指的是积的辐角的集合等于原来两个复数的辐角集合中各自任取一个，求和角，所有和角组成的集合．而积的辐角主值不一定等于这两个复数的辐角主 值的和．如－＝π，－＝π，－－＝＝π≠π＋π．arg(i)arg(1)arg[(i)(1)]argi 32232 arg(z1·z2)与argz1、argz2的关系是 arg(z1·z2)＝argz1＋argz2＋2k π(k 取某一整数) 其中整数k 使argz1＋argz2＋2k π∈[0，2π)． 3．根据三角形式的乘法法则，结合向量知识，可以对复数乘法的几何意义解释如下： 在复平面内作出z1、z2对应的向量，将向量按逆时针方向旋转一个角θ2(若θ2＜0，则 按顺时针方向旋转一个角|θ2|)，再把它的模变为原来的r2倍，所得向量 就表示积z1z2． 也就是说，复数乘法实质上就是向量的旋转和伸缩．旋转方向与角度取决于从另一复数的辐角集中取出来的值，伸长或缩短及其倍数取决于另一复数的模的大小． 4．将两个复数相乘的结果推广到有限个复数相乘，即为 r1(cos θ1＋isin θ1)·r2(cos θ2＋isin θ2)·…·rn(cos θn ＋isin θn) ＝r1r2…rn[cos(θ1＋θ2＋…＋θn)＋isin(θ1＋θ2＋…＋θn)](n ≥2)． 可以用数学归纳法说明： 1°当n ＝2时，乘法公式成立．
课题：4．2复数的运算（一） 教学目的：掌握复数的加法运算及意义 教学重点：复数加法运算． 教学难点：复数加法运算的运算率 授课类型：新授课 课时安排：1课时 教具：多媒体、实物投影仪 教学过程： 一、复习引入： 1.虚数单位i:(1)它的平方等于-1，即21 i=-; (2)实数可以与它进行四则运算，进行四则运算时，原有加、乘运算律仍然成立 2. i与－1的关系: i就是－1的一个平方根，即方程x2=－1的一个根，方程x2=－1的另一个根是－i 3. i的周期性：i4n+1=i, i4n+2=-1, i4n+3=-i, i4n=1 4.复数的定义：形如(,) +∈的数叫复数，a叫复数的实部，b叫复 a bi a b R 数的虚部全体复数所成的集合叫做复数集，用字母C表示* 3. 复数的代数形式:复数通常用字母z表示，即(,) =+∈，把复 z a bi a b R 数表示成a+bi的形式，叫做复数的代数形式 4. 复数与实数、虚数、纯虚数及0的关系：对于复数(,) +∈，当 a bi a b R 且仅当b=0时，复数a+bi(a、b∈R)是实数a；当b≠0时，复数z=a+bi叫做虚数；当a=0且b≠0时，z=bi叫做纯虚数；当且仅当a=b=0时，z就是实数0. 5.复数集与其它数集之间的关系：N Z Q R C. 6. 两个复数相等的定义：如果两个复数的实部和虚部分别相等，那么我们就说这两个复数相等即：如果a，b，c，d∈R，那么a+bi=c+di?a=c，b=d 一般地，两个复数只能说相等或不相等，而不能比较大小.如果两个复数都 是实数，就可以比较大小只有当两个复数不全是实数时才不能比较大小 7. 复平面、实轴、虚轴： ∈R)可用点Z(a，b)表示，这个建立了直角坐标系来表 示复数的平面叫做复平面，也叫高斯平面，x轴叫做实 轴，y轴叫做虚轴 实轴上的点都表示实数 对于虚轴上的点要除原点外，因为原点对应的有序
高中数学《复数》练习题 一．基本知识：复数的基本概念 （1）形如a + b i 的数叫做复数（其中R b a ∈，）；复数的单位为i ，它的平方等于－1，即1i 2-=.其中a 叫做复数的实部，b 叫做虚部 实数：当b = 0时复数a + b i 为实数 虚数：当0≠b 时的复数a + b i 为虚数； 纯虚数：当a = 0且0≠b 时的复数a + b i 为纯虚数 （2）两个复数相等的定义： 00==?=+∈==?+=+b a bi a R d c b a d b c a di c bi a ）特别地，，，，（其中，且 （3）共轭复数：z a bi =+的共轭记作z a bi =-； （4）复平面： z a bi =+，对应点坐标为(),p a b ；（象限的复习） （5）复数的模：对于复数z a bi =+，把z =z 的模； 二．复数的基本运算：设111z a b i =+，222z a b i =+ （1） 加法：()()121212z z a a b b i +=+++； （2） 减法：()()121212z z a a b b i -=-+-； （3） 乘法：()()1212122112z z a a b b a b a b i ?=-++ 特别22z z a b ?=+。 （4）幂运算：1i i =21i =-3i i =-41i =5i i =61i =-?????? 三．复数的化简 c di z a bi +=+（,a b 是均不为0的实数）；的化简就是通过分母实数化的方法将分母化为实数：()()22 ac bd ad bc i c di c di a bi z a bi a bi a bi a b ++-++-==?=++-+ 四．例题分析 【例1】已知()14z a b i =++-，求 （1）当,a b 为何值时z 为实数（2）当,a b 为何值时z 为纯虚数 （3）当,a b 为何值时z 为虚数（4）当,a b 满足什么条件时z 对应的点在复平面内的第二象限。 【变式1】若复数2(1)(1)z x x i =-+-为纯虚数，则实数x 的值为 A ．1- B ．0 C 1 D ．1-或1 【例2】已知134z i =+；()()234z a b i =-+-，求当,a b 为何值时12=z z 【例3】已知1z i =-，求z ，z z ?；
英语单词复数形式的规律 1、名词由单数变复数的基本方法如下： ①在单数名词词尾加s。如：map → maps,boy→ boys,horse→ horses, table→ tables. ②s,o,x ,sh,ch结尾的词加es.如：class→classes, box→boxes, hero→heroes, dish→dishes, bench→benches. [注]：少数以o结尾的词,变复数时只加s。如：photo→photos, piano→pianos. ③以辅音字母加y结尾的名词,变y为i,再加es。如：family→families, city→cities, party→parties. ④以f或fe结尾的名词,变f或fe为v,再加es。如：shelf→shelves, wolf→wolves, life→lives, knife→knives. 2、不规则变化：man→men, woman→women, sheep→sheep,tooth→teeth, fish→fish, child→children, ox→oxen, goose→geese 不可数名词一般没有复数形式,说明其数量时,要用有关计量名词。如：a bag of rice→two bags of rice, a piece of paper→three pieces of paper, a bottle of milk→five bottles of milk. 动名词的规则变化 1 一般情况下，直接在动词后加－ing （现在进行时） eg: work ---- working study ----- studying 2 动词以不发音的－e结尾，要去－e加－ing take ----- taking make ----- making 3 重读闭音节的动词，要双写词尾字母，再 加－ing cut ----- cutting put begin 4 以－ie结尾的动词，把变成y再加－ing lie ----- lying tie ----- tying die ----- dying