# Simple Complex number problem,

1. Oct 12, 2005

### UnD

just having a problem with these 3 questions.
z E C such that Im z=2 and z^2 is real find z
well from my knowledge, it's will be x+2i, z^2 is (x^2 -4) + 4xi, since Im z= 2
then 4xi= 2? doesn't it, Umm don't really know what to do next

2nd questions
z E C such that Re z= 2Im z, and z^2 -4i is real, find x.
Umm don't have a clue on this one

Also 1 more
z EC such that z/(z-i) is real, show that z is imaginary

Thanks very much if you can help

2. Oct 12, 2005

### HallsofIvy

Staff Emeritus
No, it's the imaginary part of z that is 2, not z2. Knowing that z2 is real tells you that Im(z2)= 0.

z= 2x+ xi= x(2+ i). z2= x2(4+ 2i- 1)= x2(3+ 2i) so z2- 4i= 3x2+ (2x2- 4)i.
If z2- 4i is real, then the imaginary part of that is 0.

The standard way of dealing with fractions is to multiply numerator and denominator by the complex conjugate of the denominator. Here that would be z*+ i (z* being the complex conjugate of z). Remembering that z(z*)= |z|2, a real number, what condition on z makes z/(z-i) real?

Or, again, write z= x+iy so that z*+i= x- yi+ i= x+ (1-y)i.

3. Oct 12, 2005

### UnD

Thanks, Still having a bit of problem with question 1 and 2. If you could explain again, It would be great.

4. Oct 12, 2005

### TD

For 1: start with the general z = x+iy. Taking the imaginary part and lettting it equal 2 will solve one of the two unknowns directly. Then z², take the imaginary part of it, let it equal 0 this time, since it has to be purely real.

Use a similar strategy for 2, but HallsofIvy already did most of the work for you. What part don't you understand?