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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2I could not think of anything else..sry – malloc Oct 12 '13 at 10:54
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I didn't mean to be mean, I forgot to put the :) at the end, sorry – jimjim Oct 12 '13 at 10:57
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1don't worry, It's nothing! :) – malloc Oct 12 '13 at 11:32
3 Answers
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Hint: first of all, set $w=z^2$, so the equation simplifies to $$ \bar{w}^2+w=16i $$
egreg
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I can apply the resolutive formula for the equations of the second degree? – malloc Oct 12 '13 at 11:17
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@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 $z$. – egreg Oct 12 '13 at 11:19
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@malloc Sorry, but this problem requires solving a quartic that has no “simple” roots. – egreg Oct 12 '13 at 12:47
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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 $b$.
Shobhit
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@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. – Shobhit Oct 12 '13 at 11:27