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I was trying to solve this equation: $$(\bar{z})^4+z^2=16i$$

but do not know where to start, I tried to carry out the powers, but then I do not know to continue, in my book there is not enough information. where do I start?

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    $\begingroup$ I could not think of anything else..sry $\endgroup$
    – malloc
    Oct 12, 2013 at 10:54
  • $\begingroup$ I didn't mean to be mean, I forgot to put the :) at the end, sorry $\endgroup$
    – jimjim
    Oct 12, 2013 at 10:57
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    $\begingroup$ don't worry, It's nothing! :) $\endgroup$
    – malloc
    Oct 12, 2013 at 11:32

3 Answers 3

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HINT:If $z=a+bi$ then $z^*=a-bi$ $$(a-bi)^4+(a+bi)^2=16i$$

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Hint: first of all, set $w=z^2$, so the equation simplifies to $$ \bar{w}^2+w=16i $$

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  • $\begingroup$ I can apply the resolutive formula for the equations of the second degree? $\endgroup$
    – malloc
    Oct 12, 2013 at 11:17
  • $\begingroup$ @malloc No, but you lower the degree; when you've found the value(s) for $w$, just get their square roots to get the solutions for $z$. $\endgroup$
    – egreg
    Oct 12, 2013 at 11:19
  • $\begingroup$ so $(\bar{w})=+-sqrt(-w+16i)$ $\endgroup$
    – malloc
    Oct 12, 2013 at 11:24
  • $\begingroup$ the result with wolfram alpha is huge, there are a lot of radicals $\endgroup$
    – malloc
    Oct 12, 2013 at 11:58
  • $\begingroup$ @malloc Indeed; I think this is not a good exercise. $\endgroup$
    – egreg
    Oct 12, 2013 at 12:22
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HINT:

Let $z=a+ib$ and $\bar{z}=a-ib$

$(\bar{z})^4=(a-ib)^4$ and $z^2=(a+ib)^2$ evaluate $(\bar{z})^4$ and $z^2$, add them and equate them to $16i$.

equate real parts to $0$ and imaginary parts to $16$ and solve for $a$ and $b$.

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  • $\begingroup$ This is copy of my answer. Have you something new? $\endgroup$
    – Adi Dani
    Oct 12, 2013 at 11:25
  • $\begingroup$ @AdiDani hell no,u just said evaluate L.H.S, the OP already knows that he don't know what to do next, i just gave him an idea. Have a look at the question again. $\endgroup$
    – Shobhit
    Oct 12, 2013 at 11:27

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