# 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?

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May be short route be to note that $Z=0$ satisfies the equation and then any $z\ne0$ can be expressed as $re^{i\theta}$ and the equation gives $r=3$ and no restriction $\theta$. – Tapu Nov 29 '11 at 18:58
Now that you've seen two solutions to your first question, why not show us how far you can get toward a solution of the equation in your edit? – Gerry Myerson Nov 29 '11 at 22:48

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 $\theta$.

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In steps:

1. $z\cdot \bar z = |z|^2$;

2. $|3\cdot z| = 3|z|$;

3. $|z|^2 = 3|z|\Leftrightarrow |z| = 0\text{ or }|z| = 3\Leftrightarrow z = 0\text{ or }z = 3\mathrm e^{i\phi}$ for $\phi\in\mathbb R$.

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