Equations in the complex field

Problem

Find all complex solutions: $a)\ z^{3}+6z^{2}+25z=0\,;\quad b)\ -4iz^{4}-4z^{3}+4z-4i=0\,;\quad c)\ 4^{-iz}=-2\,;\quad d)\ \ln(-z)=1-2i.$

Solution

a) $z(z^{2}+6z+25)=0\Rightarrow z=0$ or $z=\dfrac{-6\pm\sqrt{36-100}}{2}=-3\pm4i$. b) Dividing by $-4$: $iz^{4}+z^{3}-z+i=0$. Since $iz+1=i(z-i)$, it factors as $(z-i)(iz^{3}-1)=0$: $z=i$ or $z^{3}=-i$, i.e. $z=e^{-i\pi/6},\,e^{i\pi/2},\,e^{i7\pi/6}$. The solutions are $z=i$ (double), $z=\tfrac{\sqrt3}{2}-\tfrac12 i$, $z=-\tfrac{\sqrt3}{2}-\tfrac12 i$. c) $-iz\ln4=\ln 2+i\pi(1+2k)\Rightarrow z=-\dfrac{\pi(1+2k)}{2\ln2}+\dfrac12 i$; e.g. $z\approx\mp2{,}266+0{,}5i$. d) $-z=e^{1-2i}=e(\cos 2-i\sin 2)\Rightarrow z=1{,}131+2{,}472\,i$. $\boxed{\,a)\ 0,\,-3\pm4i\quad b)\ i^{(2)},\,\pm\tfrac{\sqrt3}{2}-\tfrac12 i\quad c)\ z=-\tfrac{\pi(1+2k)}{2\ln2}+\tfrac12 i\quad d)\ 1{,}131+2{,}472i\,}$