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let $z = 1 +i$

Find all complex solutions such that $z^2 + \bar z^2 = 0$.

My working out:

$z^2 = -\bar z^2 = -(1-i)^2 = 2i$

so $z^2 = 2i$

hence $r^2 = 2 \implies r = \sqrt 2$

mod: $2\theta = \frac{\pi}{2} + 2k\pi \implies \theta = \frac{\pi}{4} + k(\pi)$ where $k = 0, 1$

overall roots are $z = \sqrt 2 \operatorname{cis} \left(\frac\pi4 +k\pi\right)$

Is my working out and solution correct?

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I think question should be either proof or derivation of the roots as if $z=a+ib, z^2=a^2-b^2+2ab\cdot i$ $\implies \bar z=a-ib,(\bar z)^2=(a-ib)^2=a^2-b^2-2ab\cdot i$ $\implies z^2+(\bar z)^2=2(a^2-b^2)=0$ if $a=\pm b,$ like here – lab bhattacharjee Jun 9 '13 at 14:37
As mentioned by Lab, that's what I got too. All complex number on the lines y = x and y = -x work – imranfat Jun 9 '13 at 14:41
auriga123: check the edit (click edit history) to view the formatting so you can learn how to format in "mathjax". But look over the post to ensure I formatted your post correctly. – amWhy Jun 9 '13 at 14:42
If you start with "Let $z=1+i$", then what is the unknown you are searching solutions for? – Hagen von Eitzen Jun 9 '13 at 14:42
understood the problem...thanks! – auriga123 Jun 10 '13 at 11:58

You're working seems a bit confused. It's like you're verifying a solution and trying to solve for a solution at the same time. Do one or the other. Either substitute in $z=1+i$ and show that this satisfies the equation (no solving involved!) or substitute in $z=a+bi$ and solve to find the conditions on $z$ to satisfy the equation.

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