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I need to solve for z in equation below.

${{z}^{2} = \bar{z}}$

If I can substitute ${\bar{z}}$ in terms of $ {z}$, then I can go about solving this equation.

Question

What would be this substitute expression?

UPDATE

There are four roots for this equation given at back of the book and they are ${0,1, -\frac{1}{2} + \frac{\sqrt3}{2}i,-\frac{1}{2} - \frac{\sqrt3}{2}i }$.

I can see how to get ${0,1}$, but confused about the last two answers in above list. I can assume ${z^{2}}$ is a pure real number since only for real numbers the complex number equals it's conjugate, and this will lead me to the first two roots.

Sunil
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5 Answers5

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We don't need that

Let $z=a+ib$ where $a,b$ are real.

$\implies a^2-b^2+i(2ab)=a-ib$

Now equate the real & the imaginary parts.

lab bhattacharjee
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  • Thanks. I will try this out. – Sunil Feb 12 '17 at 13:28
  • This results in a longer solution but it gives the answers. However, i need to check each time I get a root. Is it the right approach to check the root each time I get a root? – Sunil Feb 12 '17 at 13:35
  • @Sunil, You have to validate the roots in neither of my answers. http://math.stackexchange.com/questions/55445/when-do-we-get-extraneous-roots – lab bhattacharjee Feb 12 '17 at 13:37
  • So if we did not square both side or cubed both sides or changed the equation by introducing a higher power, and just took the original equation then there will no extraneous roots? – Sunil Feb 12 '17 at 13:39
  • @Sunil, See the answers. – lab bhattacharjee Feb 12 '17 at 13:40
  • Thanks for your help. I saw the answer, but when the max power of ${z}$ is increased in an equation by some operation then definitely one must check for extraneous root since the resulting equation is of higher degree compared to original equation. A higher degree equation means extraneous roots. – Sunil Feb 12 '17 at 14:05
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Multiply both sides by $z$ to obtain $$z^3=\bar z z=|z|^2 $$ In particular, $z^3\in\Bbb R$. Now apply modulus to the above to find $|z|^3=|z|^2$, i.e., $|z|=0$ or $|z|=1$. Hence you need only check $z=0$ and the three solutions of $z^3=1$.

Hagen von Eitzen
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  • Thanks for your nice tip. So there will be 4 roots using your approach. I will try it out and let you know. – Sunil Feb 12 '17 at 13:30
  • Thanks for your excellent tip. It also works and gives the same answers as mentioned in book. – Sunil Feb 12 '17 at 13:42
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As $|z|=|\overline z|$

Taking modulus we have $|z|^2=|\overline z|\iff|z|(|z|-1)=0$

If $|z|\ne0, |z|=1$ let $z=e^{it}$ where $t$ is real.

So, we have $(e^{it})^2=e^{-it}\iff e^{3it}=1=e^{2m\pi i}$ where $m$ is any integer

$t=2m\pi/3$

Alternatively, start with $z=re^{it}$

lab bhattacharjee
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It seems to me that you are trying to solve the following $$z^2=\bar{z}$$ and not the following $$|z|^2=\bar{z}\,.$$ In that case, replacing $\bar{z}$ with a function of $z$ (such as, $\Re(z)+i \Im(z)$, where $\Re$ and $\Im$ denotes the real and imaginary part of $z$) seems overly complicated to me.

Instead, I would suggest to replace $z$ with $x+i\,y$, where $x=\Re(z)$ and $y=\Im(z)$. These replacements will yield the following $$x^2-y^2+i \,2\, x\,y = x - i\,y\, .$$ Solving for the imaginary part yields the following $$ 2\,x\,y = -y \implies x = -\frac12 \text{ or } y=0$$ Replacing $x=-\frac12$ this into the real part yields the following $$ \frac14-y^2 = -\frac12 \implies y=\pm \sqrt{\frac34}$$ Replacing $y=0$ this into the real part yields the following $$ x^2 = x \implies x\in\{0,1\}$$ Putting all together yields the following four roots for the equation $z^2=\bar{z}$ $$z\in\Big\{0,1,-\frac12\pm\sqrt{\frac34}\Big\}.$$

Hope this helps.

HGE
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Given $z^{2}=\bar {z} $ , this implies $ |z|^{2}=|\bar{z}|$ or, $ z \bar {z}=|\bar{z}| $ or, $ |z|^{2}=|z| $. Then , $ (|z|-1)|z|=0$ , implies $ either \ |z|-1=0 , \ or \ |z|=0 $.$ \begin {align}Then \ either \ z=1,or, z=i, \ or, -i\ or, \ or \ z=0 \end{align} $

Arifa zaman
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