13 2016 - 2017 H2 Maths Complex Numbers Lecture Questions (Stu)

National Junior College Mathematics Department 2017 Complex Numbers Page 1 of 14 National Junior College 2016 – 2017 H2 Mathematics Complex Numbers Lecture Questions Key questions to answer:  What is a complex number? o How is the complex number i defined? o How do we write a complex number in its Cartesian form? o What do we mean when we say that two complex numbers are equal? Lecture Reading: Section 1 Definition (Imaginary Unit) The imaginary unit, i, is a number such that 2i1 . Hence, i1 . Definition (Complex Number in Cartesian form) A complex number, z , is a number of the form ixy , where ,xy  . Question 1 (Real and Imaginary Parts of a Complex Number) Write down the real and imaginary parts of the following complex numbers: (a) 3i (b) 2i (c) 7 (d) i6 Solution: (a)    Re 3 i 3, Im 3 i 1    (b)    Re 2i 0, Im 2i 2    (c) (d)    Re 7 7, Im 7 0    Re i 6 6, Im i 6 1    Note: 1. The imaginary part of a complex number does not include ‘i'. 2. A complex number having only the imaginary part (i.e. real part equals zero) is known as a purely imaginary number.

National Junior College Mathematics Department 2017 Complex Numbers Page 2 of 14 Question 2 (Equality of Complex Numbers) Find x and y, where ,xy  , if  2 3i 2i 1 ix y x y     . Solution:      2 3i 2i 1 i 2 3i 2 ix y x y x y x x y           Comparing real part on both sides, we get 2 0 (1)x y x x y      Comparing imaginary part on both sides, we get 3 2 (2)xy   Solving equations (1) and (2) simultaneously, we get 1x and 1y . Alternatively, we may group all the real and imaginary parts respectively on one side of the equation, giving us    2 2 3 2 i 0x y x y     . Comparing real and imaginary parts on both sides, we get the equations 2 2 0 and 3 2 0x y x y     , which we can solve simultaneously. Question 3 Given that i , where , ,z x y x y   f ind the value of x such that  2 i i 1 .x z z x    Identify the mistake in the following proposed solution and correct it. Comparing real part on both sides: 2xz Comparing imaginary part on both sides: 1zx  (*) Substitute 2zx into equation (*): 121 3x x x    . Correct solution:        2 i i i i 1 2 i i 1x x y x y x x y x x y x             Comparing real part on both sides: 2x y x x y    Comparing imaginary part on both sides: 1 2 1x y x x y      Solving both equations simultaneously, we get 1x .

National Junior College Mathematics Department 2017 Complex Numbers Page 3 of 14 Key questions to answer:  How do we carry out arithmetic operations (addition, subtraction, multiplication, division and taking square root) on complex numbers in Cartesian form? o How do we relate the addition and subtraction of complex numbers to addition and subtraction of vectors? Prerequisite knowledge: Complex number in Cartesian form ; Vector addition and subtraction. Lecture Reading: Section 2 Question 4 (Arithmetic Operations on Complex Numbers) Express the following complex numbers in the form i , where ,x y x y . (a) (3 + 3i) − (1 − 2i) (b) (2 + 4i)(2 – i) (c) 2 4i 2i   (d) 3 i Solution: (a) (3 + 3i) − (1 − 2i) = 3 − 1 + 3i + 2i = 2 + 5i (b) (2 + 4i)(2 – i) = 2(2 – i) + 4i(2 – i) = 4 – 2i + 8i – 4i2 = 4 + 6i + 4 = 8 + 6i (c) 2 4i 2 4i 2 i 8 6i 8 6.i2 i 2 i 2 i 4 1 5 5           Recall:    22a b a b a b    for ,.ab  In fact, this property is true for ,.ab  Thus, in (c),     222 i 2 i 2 i 4 1 5.        (d) 2 3 3 i 3i 3ii i i i    Note: 1. In (c), 2i is the complex conjugate of 2i , which we will further discuss in Question 7. Observe that the product of a complex number and its conjugate gives us a real number. 2. It is useful to know (and remember) that 1 ii  , as seen in (d).

National Junior College Mathematics Department 2017 Complex Numbers Page 4 of 14 Question 5 (Solving Simultaneous Equations) Solve the simultaneous equations 2 i 2pq and 2 8 2i 0pq    , for the complex numbers p and q in Cartesian form. Solution: 2 i 2pq ---- (1) 22 8 2i 0 8 2ip q q p        ---- (2) Substitute (2) into (1):     2 22 8 2i i 2 i 2 8i 0 2 4 4i 8i 2i 2 4 32 2i 2 6 2 6 or 2i 2i 4i or 2i pp pp p p pp pp                 When  2 4i, 4i 8 2i 8 2i.pq       When  2 2i, 2i 8 2i 4 2i.pq      Question 6 (Square Roots of Complex Numbers) Find the square roots of 3 4i . Solution: The question requires us to evaluate 3 4i . Let 3 4i i xy   . Then   2 22i 3 4i 2i 3 4ix y x y xy        . Comparing the real and imaginary parts on both sides, we have: 22 3 ----- (1)xy ; 2 4 2 ----- (2)xy xy   Solving simultaneously (write out the workings yourself), we have 2x , 1y or 2x , 1y . The square roots of 3 − 4i are 2i and 2i .

National Junior College Mathematics Department 2017 Complex Numbers Page 5 of 14 Can we answer Question 6 using the GC? Partly. Observed that the GC gives us only one of the two square roots that we obtained through an algebraic method. Key questions to answer:  How do we find the conjugate of a complex number?  What are the properties of complex conjugates and their applications? Prerequisite knowledge: Arithmetic operations on complex numbers Lecture Readings: Section 3 Definition (Complex Conjugate) The complex conjugate of a complex number iz x y is defined to be the complex number ixy and is denoted by *z . Question 7 (Properties of Complex Conjugates) If 1 2iz  and 2 1 3iz  , find (a) * 11zz (b) * 11zz (c) * 11zz (d)   * 12zz (e) * 12 2 2 1 zz z     . Solution: (a)        * 1 1 1 2 i 2 i 4 2Rez z z       (b)        * 1 1 1 2 i 2 i 2i 2i Imz z z       (c)       *2 11 2 i 2 i 2 2 i i 2 i 4 2i 2i i 5zz             (d)      * ** 1 2 1 2 2 i 1 3i 5 5iz z z z      Compare this with         *** 12 2 i 1+3i 5 5i 5 5izz       . (e)           * ** 1 2 1 2 * 22 22 2 2 i 1 3i 5 i 5 i i 1 1 5i1 1 1 3i 1 3i 3 3 z z z z zz                  Note: 1. In general,  * 2Rez z z and  * 2i Imz z z , as seen in (a) and (b). 2. The conjugate of a real number is the number itself. Let’s be intellectually careful.

National Junior College Mathematics Department 2017 Complex Numbers Page 6 of 14 Question 8 (Properties of Complex Conjugates) Using Binomial Theorem, expand  5 2 3i in the form i, where , .a b a b Hence, simplify  5 2 3i in a similar form. Solution:                5 2 3 4 55 4 3 22 3i 2 5 2 3i 10 2 3i 10 2 3i 5 2 3i 3i 32 240i 720 1080i 810 243i =122 597i                       55* *5 * 2 3i 2 3i 2 3i 122 597i 122 597i        Note: In general,     * * nnzz  , where n . Can we use the GC to perform these arithmetic operations on complex numbers? Yes, unless the question specifies that no calculator is allowed or an exact answer (or specific method) is required. Refer to Appendix I of the lecture notes on some basics of using GC in complex numbers. Key questions to answer:  What can we say about the roots of polynomial equations with real coefficients? - How do we solve polynomial equations with real coefficients? - Understand that complex roots of a polynomial equation with real coefficients occur in conjugate pairs