COMPLEX NUMBERS - Need Help?

Question

I am having trouble with complex #'s .

I have three problems.. for one of them I have the answer ; however I don't know how to do the work for any of them.


#1 .
Problem:
(2 - 3i)/(3 + 4i)

Answer: -6/25 - (17/25)i


Problem 2.
Let z = a + bi represent a general complex number. As noted in the lesson, the conjugate of z, abbreviated conj(z) or conj(a + bi) is the complex number a-bi. Also, the modulus of z, modulus(z) is the "size" of z, or SQRT(a^2 + b^2). Which of the following is true for all complex numbers?
Answer Options:

All of the following
z + conj(z) = 2a
z - conj(z) = 2bi
None of the above


Problem # 3:
Which number is the largest?
A3i
B-7
C4
D -2i
E 4 + 4i
F None of the above

I thought it was 4 but I was wrong...

Answer 1

PROBLEM 1:
(2 - 3i)
---------
(3 + 4i)

To start, multiply the numerator and denominator by the denominator's conjugate (3 - 4i)... this will get the complex number out of the denominator:

(2 - 3i) (3 - 4i)
--------------------
(3 + 4i) (3 - 4i)

Multiply out the denominator using FOIL:
(2 - 3i) (3 - 4i)
---------------------------
(3² - 12i + 12i - (4i)²)

Notice how the 12i terms cancel out:
(2 - 3i) (3 - 4i)
---------------------------
(3² - (4i)²)

Also notice how you have i² in the denominator, which you can replace with -1
(2 - 3i) (3 - 4i)
-------------------
(3² - (4)²(i²))

(2 - 3i) (3 - 4i)
-------------------
(3² - (-1)(4)²)

(2 - 3i) (3 - 4i)
-------------------
(9 - (-16))

(2 - 3i) (3 - 4i)
-------------------
25

Now multiply out the numerator:
(2*3 -8i - 9i +12i²)
------------------------
25

Replace i² with -1 again, and combine the i terms:
(6 -17i -12)
----------------
25

-6 -17i
----------
25

Finally just split it into the real and imaginary parts:
-6/25 - (17/25)i

PROBLEM 2:
z = a + bi
conj(z) = a - bi

z + conj(z)
= a + bi + a - bi
= 2a

z - conj(z)
= a + bi - (a - bi)
= a + bi - a + bi
= 2bi

So the answer is 'All of the following'

PROBLEM 3:
z = a + bi
modulus(z) is the "size" of z
modulus(z) = sqrt(a² + b²)

So let's try them:
A) 3i --> a = 0, b = 3
sqrt(0² + 3²) = sqrt(9) = 3

B) -7 --> a = 0, b = -7
sqrt(0² + (-7)²) = sqrt(49) = 7

C) 4 --> a = 4, b = 0
sqrt(4² + 0²) = sqrt(16) = 4

D) -2i --> a = 0, b = -2
sqrt(0² + (-2)²) = sqrt(4) = 2

E) 4 + 4i --> a = 4, b = -4
sqrt(4² + (-4)²) = sqrt(16 + 16) = sqrt(32) ≈ 5.66

So the answer is B has the largest "size" with modulus(z) = 7

Answer 2

(2-3i) 3-4i
--------- * ------- = (multiply by conjugate)
(3+4i) 3-4i

6-12 -8i-9i
------------------- =
9+16+12i-12i

-6-17i
--------- =
25

-6 17
---- - --- i
25 25