Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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
1  
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

2 Answers 2

up vote 3 down vote accepted

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 $\theta$.

share|improve this answer

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 $\phi\in\mathbb R$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.