Find all complex numbers $z$ in Cartesian form such that $(z-4)^4=-1$.
This is what I did: I squared root both side making $-1$ into $i$ and $(z-4)^2$ and I took square of both side again and had $i^2$ which is equal to $-1$ and the other side $(z-1)$. At the end I had $z=0$.
On
You'd better use the exponential form of $z-4=r\mathrm e^{i\theta}\enspace(r\ge 0)$. This yields the equation $$(z-4)^4=r^4\mathrm e^{4i\theta}=-1=\mathrm e^{i\pi}\iff\begin{cases} r^4=1\\ 4\theta\equiv \pi\bmod 2\pi \end{cases}\iff\begin{cases} r=1\\ \theta\equiv \frac\pi4\bmod \frac\pi2 \end{cases}$$ Can you end the computations?
On
Write the equation as $(z-4)^4=-1 = e^{i\pi+i2\pi k}$ with integer $k$. There are four solutions, given by,
$$z-4 = e^{\frac14 (i\pi+i2\pi k) },\>\>\>\>\>k=0,1,2,3$$
Thus, in Cartesian form, the four solutions are,
$$z = 4 + \cos\left[\frac\pi4(1+2k)\right]+i\sin\left[\frac\pi4(1+2k)\right] $$
with $k=0,1,2$ and $3$
This is on the right track. Except, when you square root both sides, you need to do $\pm$. So instead of getting $i$ you get $\pm i$: $$ (z - 4)^2 = \pm i $$
The next step is to take the square root again (not squaring both sides), is that what you meant to do? This part is a little tricky: when we take square root of both sides we need to know what are all the square roots of $i$ and $-i$. The square roots of $i$ are $$ \pm \left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i \right) $$ and the square roots of $-i$ are $$ \pm \left( \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} i \right) $$ So, when we take the square root of the equation $(z - 4)^2 = \pm i$, we get four possibilities (the two square roots of $i$, and the two square roots of $-i$): \begin{align*} z - 4 &= \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i \\ z - 4 &= \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} i \\ z - 4 &= -\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i \\ z - 4 &= -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} i. \end{align*} Can you find the answer in all of the four cases?