1. ## COmplex number uestion

Hi again,

I need help with this question,

Find the complex number z that satisfies the equation 50/z* - 10/z = 2 + 9i given | z | = 2sqrt(10).

I had put 50 and 10 to the same denominator ie. zz* which then gives me |z|^2, and then I subbed in the value for |z| so my denominator was 40.
I then brought 40 over to RHS and got 80+360i.
The problem comes with the LHS, I am not sure how to deal with 50z-10z*..
I thought since z+ is a conjugate, I can turn it into z by changing the sign thus 50z+10z,
but I do not think I'm right, the answer key proved me wrong anyway.

Thank you so much!
J.

2. ## Re: COmplex number uestion

what we are given:

|z| = 2√(10)

50/z* - 10/z = 2 + 9i

what i like to do is get everything over a common denominator:

50z/(zz*) - 10z*/(zz*) = 2 + 9i

50z - 10z* = (2 + 9i)(zz*)

now zz* = |z|2 = (2√(10))2 = 40 so we have:

50z - 10z* = (2 + 9i)(40)

take out a common factor of 10:

5z - z* = 8 + 36i

let's write this in terms of a and b, where z = a + bi

5(a + bi) - (a - bi) = 8 + 36i

(5a - a) + (5b + b)i = 8 + 36i

4a + 6bi = 8 + 36i

compare real and imaginary parts:

4a = 8 ---> a = 2
6b = 36 ---> b = 6.

so z = 2 + 6i. does this work?

first, we check |z| = √(a2 + b2) = √(4 + 36) = √40 = √4√(10) = 2√(10). check.

next, let's find 1/z = 1/(2 + 6i) = (2 - 6i)/(4 + 36) = (1/40)(2 - 6i)

= (1/20)(1 - 3i).

finally, let's find 1/z* = z/|z|2 = (1/40)(2 + 6i) = (1/20)(1 + 3i).

so 50/z* = (50/20)(1 + 3i) = (5/2)(1 + 3i), and 10/z = (10/20)(1 - 3i) = (1/2)(1 - 3i).

subtract the second from the first: (5/2)(1 + 3i) - (1/2)(1 - 3i) =

(5/2 - 1/2) + (15/2 + 3/2)i = 4/2 + (18/2)i = 2 + 9i. check.

3. ## Re: COmplex number uestion

Thank you!

How did you get from
1/(2 + 6i) to (2 - 6i)/(4 + 36) ?

J.

4. ## Re: COmplex number uestion

Originally Posted by Tutu
Thank you!

How did you get from
1/(2 + 6i) to (2 - 6i)/(4 + 36) ?

J.
Multiply top and bottom by the bottom's conjugate...

5. ## Re: COmplex number uestion

Originally Posted by Tutu
Thank you!

How did you get from
1/(2 + 6i) to (2 - 6i)/(4 + 36) ?

J.

for any complex number a+bi:

1/(a+bi) = (a-bi)/(a2+b2).

how do we prove this?

suppose xy = 1. then y = 1/x. so if we want to show y = 1/x, we can instead show xy = 1.

so let's prove (a+bi)[(a-bi)/(a2+b2)] = 1

the denominator in the second factor is just a real number, so we can "pull it out front":

(a+bi)[(a-bi)/(a2+b2)] = (1/(a2+b2))[(a+bi)(a-bi)]

= (1/(a2+b2))(a2-abi+abi-(bi)2)

= (1/(a2+b2))(a2-b2i2)

= ((1/(a2+b2))(a2-b2(-1))

= (1/(a2+b2))(a2+b2) = 1.

a shorter proof:

zz* = |z|2 so

zz*/|z|2 = 1, that is:

z(z*/|z|2) = 1, so 1/z = z*/|z|2.