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Homework: Complex Numbers

Module by: Kenny M. Felder. E-mail the author

Summary: This module provides practice problems related to complex numbers.

Exercise 1

( 3 + 7 i ) – ( 4 + 7 i ) = (3+7i)–(4+7i)=

Exercise 2

( 5 – 3 i ) + ( 5 – 3 i ) = (5–3i)+(5–3i)=

Exercise 3

2 ( 5 – 3 i ) = 2(5–3i)=

Exercise 4

( 5 – 3 i ) ( 2 + 0 i ) = (5–3i)(2+0i)=

Exercise 5

What is the complex conjugate of ( 5 – 3 i ) (5–3i)?

Exercise 6

What do you get when you multiply ( 5 – 3 i ) (5–3i) by its complex conjugate?

Exercise 7

What is the complex conjugate of 7?

Exercise 8

What do you get when you multiply 7 by its complex conjugate?

Exercise 9

What is the complex conjugate of 2 i 2i?

Exercise 10

What do you get when you multiply 2 i 2i by its complex conjugate?

Exercise 11

What is the complex conjugate of ( a + b i ) (a+bi)?

Exercise 12

What do you get when you multiply ( a + b i ) (a+bi) by its complex conjugate?

Exercise 13

I’m thinking of a complex number zz. When I multiply it by its complex conjugate (designated as z *z*) the answer is 25.

  • a. What might z z be?
  • b. Test it, and make sure it works—that is, that ( z ) ( z * ) = 25 (z)(z*)=25!

Exercise 14

I’m thinking of a different complex number zz. When I multiply it by its complex conjugate, the answer is 3 + 2 i 3+2i.

  • a. What might z z be?
  • b. Test it, and make sure it works—that is, that ( z ) ( z * ) = 3 + 2 i (z)(z*)=3+2i!

Exercise 15

Solve for x x and y y: x 2 + 2 x 2 i + 4 y + 40 y i = 7 – 2 i x 2 +2 x 2 i+4y+40yi=7–2i

Exercise 16

Finally, a bit more exercise with rational expressions. We’re going to take one problem and solve it two different ways. The problem is 32+i−7i3+4i32+i−7i3+4i size 12{ { {3} over {2+i} } - { {7i} over {3+4i} } } {}. The final answer, of course, must be in the form a + b i a+bi.

  • a. Here is one way to solve it: the common denominator is ( 2 + i ) ( 3 + 4 i ) (2+i)(3+4i). Put both fractions over the common denominator and combine them. Then, take the resulting fraction, and simplify it into a + b i a+bi form.
  • b. Here is a completely different way to solve the same problem. Take the two fractions we are subtracting and simplify them both into a + b i a+bi form, and then subtract.
  • c. Did you get the same answer? (If not, something went wrong…) Which way was easier?

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