# solving complex numbers

• Apr 25th 2012, 08:00 AM
cellae
solving 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 z2 + 4z' + 4 = 0 where z is a complex number.
*z' being the conjugate
• Apr 25th 2012, 08:08 AM
princeps
Re: solving complex numbers
Quote:

Originally Posted by cellae
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 z2 + 4z' + 4 = 0 where z is a complex number.
*z' being the conjugate

Let $z=a+bi$ , then :

$(a^2-b^2)+2abi+4(a-bi)+4=0$

Hence :

$\begin{cases}a^2-b^2+4a+4=0 \\2ab-4b=0 \end{cases}$
• Apr 25th 2012, 08:10 AM
cellae
Re: solving complex numbers
Why, thank you !!
• Apr 25th 2012, 08:13 AM
a tutor
Re: solving complex numbers
In this case z=z*.
• Apr 25th 2012, 08:20 AM
cellae
Re: solving complex numbers
.. Could you extrapolate?
• Apr 25th 2012, 08:31 AM
a tutor
Re: solving complex numbers
Extrapolate?

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

So with Im(z)=0, z=z' and you have the solution z=2.
• Apr 25th 2012, 08:34 AM
Plato
Re: solving complex numbers
Quote:

Originally Posted by cellae
.. Could you extrapolate?

If a polynomial has real coefficients then if $z_0$ is a root then so is $\overline{z_0}$.
Thus all that needs to be done is to solve $z^2+4z+4=0~.$
• Apr 25th 2012, 08:37 AM
cellae
Re: 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 :)