# Thread: Complex number special case modulus z=1

1. ## Complex number special case modulus z=1

verify that, if z is a complex number of modulus 1(so that modulus z=1), then modulus (3z-2i)/(3+2iz) = 1

2. ## Re: Complex number special case modulus z=1

Hey Lozeee.

Hint: Try factorizing the fraction by multiplying both sides by (3-2iz)/(3-2iz) and take a look at that number.

3. ## Re: Complex number special case modulus z=1

I've tried approaching this in two different ways and i can't get it down to one :S

4. ## Re: Complex number special case modulus z=1

Was one of them what Chiro suggested? What do you get when you multiply by (3- 2iz)/(3- 2iz)?

5. ## Re: Complex number special case modulus z=1

yes i'm making an error somewhere but i can't find it because i'm getting different answers both ways, i got 5(x+yi)-12i/5 the way Chiro suggested. When I sub in z=x+yi first and then multiply by the conjugate I get 5x-12i+13yi/13-12y so I'm gone very wrong :S

7. ## Re: Complex number special case modulus z=1

(modulus 3z-2i/3+2iz) x(multiplied by) (3-2iz)/3-2iz)
9z-6i-6iz^2+4zi^2/9-6iz+6iz-4i^2z^2
5z-6i(1+z^2)/9+4z^2
then i subbed in for z=x+yi
5(x+yi)-6i(1+(x+yi)^2)/9+4(x+yi)^2
and multiplying that out i'm getting 5x+5yi-6i(1+x^2-y^2)+12xy/9+4(x+yi)^2
and i can't get past here so i'm after messing it up somewhere i believe

8. ## Re: Complex number special case modulus z=1

I'll give it a shot.

First use the theorem that z1/z2 has modulus r1/r2 with argument theta1-theta2.

For 3z-2i we get 3(x+iy) - 2i = 3x + i(3y-2).
For 3 + 2iz we get 3 + 2i(x+iy) = (3-2y) + i(2x)

We already know that x^2 + y^2 = 1 since |z| = 1. Substituting this in later and finding the modulus above gives us:

[(3x)^2 + (3y-2)^2] = 9x^2 + 9y^2 - 12y + 4
[(3-2y)^2 + (2x)^2] = 9 - 12y + 4y^2 + 4x^2

Since x^2 + y^2 = 1 we get a simplification of:

[(3x)^2 + (3y-2)^2] = 9 - 12y + 4
[(3-2y)^2 + (2x)^2] = 9 - 12y + 4

This implies that the modulus of z1/z2 = 1.