This may sound like silly question, but I've made multiple attempts and can't figure out what do do with this question:

Find all solutions to z^{2}+ 4z' + 4 = 0 where z is a complex number.

*z' being the conjugate

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- Apr 25th 2012, 07:00 AMcellaesolving complex numbers
This may sound like silly question, but I've made multiple attempts and can't figure out what do do with this question:

Find all solutions to z^{2}+ 4z' + 4 = 0 where z is a complex number.

*z' being the conjugate - Apr 25th 2012, 07:08 AMprincepsRe: solving complex numbers
- Apr 25th 2012, 07:10 AMcellaeRe: solving complex numbers
Why, thank you !!

- Apr 25th 2012, 07:13 AMa tutorRe: solving complex numbers
In this case z=z*.

- Apr 25th 2012, 07:20 AMcellaeRe: solving complex numbers
.. Could you extrapolate?

- Apr 25th 2012, 07:31 AMa tutorRe: solving complex numbers
Extrapolate?

$\displaystyle (z+2)^2=z^2+4z+4$

So with Im(z)=0, z=z' and you have the solution z=2. - Apr 25th 2012, 07:34 AMPlatoRe: solving complex numbers
- Apr 25th 2012, 07:37 AMcellaeRe: solving complex numbers
Ok I get that.

I was confused initially because I wasn't really sure how to go about the conjagate.

That being because it's the imaginary number that get's its sign swapped. So Wasn't sure if i could change the sign of a whole complex number.

Brilliant :)