Using De Moivre's theorem to solve $(z−3+2i)^4 = z^4$ What are all the solutions to: $(z−3+2i)^4 = z^4$?
I know I have to use De Moivre's theorem which states:
$$(\cos\theta + i\sin\theta)^n=\cos\theta n + i\sin\theta n$$
 A: Note that
$$a^4-b^4=(a-b)(a+b)(a-bi)(a+bi)=0,$$
where $a=z-3+2i,b=z.$ You will find three roots (your equation is cubic).
A: To solve
$$
\left(\frac{z-3+2i}{z}\right)^4=1\tag1
$$
Using De Moivre, we want to find $\theta$ so that
$$
\begin{align}
1
&=(\cos(\theta)+i\sin(\theta))^4\tag{2a}\\
&=\cos(4\theta)+i\sin(4\theta)\tag{2b}
\end{align}
$$
which is $\theta\in\left\{0,\frac\pi2,\pi,\frac{3\pi}2\right\}$
$$
\frac{z-3+2i}z=\cos\left(\tfrac{k\pi}2\right)+i\sin\left(\tfrac{k\pi}2\right)\tag3
$$
for $k\in\{1,2,3\}$ (we won't get a solution for $k=0$). That is,
$$
z=\frac{3-2i}{1-\cos\left(\frac{k\pi}2\right)-i\sin\left(\frac{k\pi}2\right)}\tag4
$$
We get the three solutions
$$
\left\{\frac{5+i}2,\frac{3-2i}2,\frac{1-5i}2\right\}\tag5
$$
A: There's not realy a need to use De Moivre's Theorem to solve this.
Let a=3-2i:
\begin{align}
   (z-a)^4=z^4 &\Leftrightarrow (z-a)^4 - z^4= 0 \\
& \Leftrightarrow \big((z-a)^2-z^2 \big)\big( (z-a)^2 + z^2\big) =0 \\
& \Leftrightarrow z^2 -2az +a^2 - z^2=0 \quad \vee \quad z^2-2az + a^2 +z^2=0\\
& \Leftrightarrow z=\frac{a}{2} \quad \vee \quad 2z^2 -2az+a^2=0  \\
\end{align}
Solving the quadratic equation
\begin{align}
   2z^2 -2az+a^2=0  & \Leftrightarrow z= \frac{2a \pm \sqrt{4a^2-8a^2}}{4}\\
& \Leftrightarrow z = \frac{2a \pm 2i\sqrt{a^2}}{4} \\
& \Leftrightarrow z=\frac{a \pm ai}{2}
\end{align}
Substituting $a$, we get the following solutions:
$$z=\frac{3}{2}-i \vee z=\frac{5}{2}+\frac{i}{2} \vee z=\frac{1}{2}-\frac{5i}{2}$$
I hope I have been of some assistance.
A: Specifically addressing your edit, $\sqrt{20 + 48i} = 2 \sqrt{5 + 12i}$.
Now if $\sqrt{5 + 12i} = a + bi$, then squaring both sides, $5 + 12i = (a^2 - b^2) + 2abi$, or that $5 = a^2 - b^2, 12 = 2ab$. Since $b = \frac{6}{a}$, you can form a quadratic with this information, which gives $a + bi = ±(3 + 2i)$.
