Finding complex solutions of an equation How does one solve this equation. I would like to see the solution of this problem in steps.
$z\cdot\bar{z}=\left|3\cdot z \right|$
EDIT: Is it possible to solve this by converting to the form $z=a+b\cdot i$
What about the solution of this equation.
$z\cdot\bar{z}-z^{2}=1-i$
EDIT2:
$a^2+b^2-(a+b\cdot i)(a+b\cdot i) = 1 - i$
$a^2+b^2-a^2-ab\cdot i - ab\cdot i + b^2=1-i$
$2b^2-2ab\cdot i = 1-i$
And we keep in mind that two imaginary numbers are equal if their real and imaginary parts are the same.
$2b^2 = 1$ and $-2ab=-1$
So $b = \pm \frac{1}{\sqrt{2}}$
and $a=\frac{1}{2b}\Rightarrow a=\pm \frac{\sqrt{2}}{2}$.
Is this correct?
 A: In steps:


*

*$z\cdot \bar z = |z|^2$;

*$|3\cdot z| = 3|z|$;

*$|z|^2 = 3|z|\Leftrightarrow |z| = 0\text{ or }|z| = 3\Leftrightarrow z = 0\text{ or }z = 3\mathrm e^{i\phi}$ for $\phi\in\mathbb R$.
A: First, note that for any complex number $z=a+bi$, we have
$$z\cdot \bar{z}=(a+bi)\cdot(a-bi)=a^2+abi-abi+b^2(i)(-i)=a^2+b^2=\left(\sqrt{a^2+b^2}\right)^2=|z|^2.$$
Now note that for any complex number $z=a+bi$ and real number $t$, we have
$$|t\cdot z|=|t(a+bi)|=|(ta)+(tb)i|=\sqrt{(ta)^2+(tb)^2}=\sqrt{(t^2)(a^2+b^2)}=$$
$$\sqrt{t^2}\sqrt{a^2+b^2}=|t|\sqrt{a^2+b^2}=|t|\cdot|z|$$
(In fact, it is true that for any two complex numbers $w$ and $z$, we have $|w\cdot z|=|w|\cdot|z|$.)
These are both important facts to know in general. 
Thus, starting from the equation
$$z\cdot \bar{z}=|3\cdot z|$$
we get
$$|z|^2=3\cdot|z|.$$
Now treat $|z|$ as a real number to be solved for - that is, think of it as if we are solving
$$x^2=3x.$$ Note that there are two solutions, i.e. two possible values for $|z|$. Do you see what they are?
Finally, note that the set of complex $z$ for which $|z|=c$ forms a circle of radius $c$ in the complex plane; using polar coordinates, i.e. $z=re^{i\theta}$, we have that $|z|=c$ if and only if $$|z|=|re^{i\theta}|=|r|\cdot|e^{i\theta}|=|r|\cdot1=|r|=r=c,$$
so the complex $z$ for which $|z|=c$ are the complex numbers of the form $ce^{i\theta}$ for some $\theta$.
