Among the exercises I was solving these are the ones I don't understand and I don't know if the process or solutions are ok, so please correct whenever I went wrong. Thanks everyone. (I'm supposed to solve by converting to trigonometric form when it's convenient and then finding roots.)
a) $z^3-z^{-3}=i$
$\frac{z^6-1}{z^{3}}=i$
$z^6-1=iz^3$
$z^3(z^3-i)=1$
1. $z^3=1$
$z=cis(\frac{2π+2kπ}{3})$, k=0,1,2
2. $z^3-i=1$
$z^3=\sqrt{2}cis(\frac{π}{4})$
$z=2^{\frac{1}{6}}cis(\frac{\frac{π}{4}+2kπ}{3})$, k=0,1,2
b) $(z-i)^6=-64$
$\sqrt{((z-i)^3)^2}=\sqrt{-64}$ square root to both sides
$(z-i)^3=\sqrt{-1*64}$
$(z-i)^3=8i$
$z-i=2cis(\frac{\frac{π}{2}+2kπ}{3})$
$z=2cis(\frac{5kπ}{6})+i=2cis(\frac{5kπ}{6})+cis(\frac{π}{2})$, k=0,1,2 ?
c) $z^5(1+i)=\bar z $ multiplying both sides by z
*$z^6(1+i)=\mid{z}\mid^2$ using absolute value on both sides
$\mid{z}\mid^6\mid{1+i}\mid=\mid{z}\mid^2$ dividing both sides by $\mid{z}\mid^2$
$\mid{z}\mid^4=\frac{1}{\mid{1+i}\mid}=\frac{1}{\sqrt{2}}$
$\mid{z}\mid=\frac{1}{2^\frac{1}{8}}$ then using * again
$z^6(1+i)=\frac{1}{2^\frac{1}{4}}$
$z^6=\frac{1}{2^\frac{1}{4}(1+i)}$
Don't know how to proceed.
I would write $$z^6-1=iz^3$$ substituting $$t=z^3$$ we get $$t^2-it-1=0$$ now you have a quadratic to solve.