# Solve: $(\bar{z})^4+z^2=16i$

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?

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

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

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

• I can apply the resolutive formula for the equations of the second degree? Oct 12, 2013 at 11:17
• @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 $z$. Oct 12, 2013 at 11:19
• so $(\bar{w})=+-sqrt(-w+16i)$ Oct 12, 2013 at 11:24
• the result with wolfram alpha is huge, there are a lot of radicals Oct 12, 2013 at 11:58
• @malloc Indeed; I think this is not a good exercise. Oct 12, 2013 at 12:22

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 $b$.

• This is copy of my answer. Have you something new? Oct 12, 2013 at 11:25
• @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. Oct 12, 2013 at 11:27