$\displaystyle \overline{z}$ denotes the conjugate of $\displaystyle z$ and is defined as
$\displaystyle \ovelrine{z}=a-bi$.

Substituting this into the equation and simplifying yields:
$\displaystyle \begin{align*} a-bi&=a+bi\\ a-bi-a-bi&=0\\ -2bi&=0\\ b&=0. \end{align*}$

So the equation is only true if $\displaystyle b=0$.

The magnitude of a complex number $\displaystyle z=re^{i\theta}$ is defined as
$\displaystyle \vert z\vert=r$.

If $\displaystyle z=e^{i\theta}$, then $\displaystyle r=1$, so $\displaystyle \vert e^{i\theta}\vert$=1.

Note that $\displaystyle \sqrt3+i$ lies in the first quadrant of the complex plane.

Any nonzero complex number $\displaystyle a+bi$ can be written in the form $\displaystyle re^{\theta i}$, where
$\displaystyle r^2=a^2+b^2$ and
$\displaystyle \tan(\theta)=\frac{b}{a}$.
(We stipulate that  $\displaystyle -\pi< \theta\leq\pi$ is in radians.)

Conversely, a nonzero complex number $\displaystyle re^{\theta i}$ can be written in the form $\displaystyle a+bi$, where
$\displaystyle a=r\cos(\theta)$ and
$\displaystyle b=r\sin(\theta)$.

We can convert $\displaystyle \sqrt3+i$ by using the formulas above:
$\displaystyle r^2=a^2+b^2=(\sqrt3)^2+1^2=3+1=4$
$\displaystyle r=\sqrt4=2$,
and
$\displaystyle \tan(\theta)=\frac{b}{a}=\frac{1}{\sqrt3}$
$\displaystyle \theta=\frac{-7\pi}{6},\frac{\pi}{6}$

Since $\displaystyle \frac{\pi}{6}$ lies in the first quadrant of the complex plane, as does $\displaystyle \sqrt3+i$, $\displaystyle \theta = \frac{\pi}{6}$.

So $\displaystyle \sqrt3+i=2e^{\frac{\pi}{6}i}$.

We now substitute this into our original expression and expand.
$\displaystyle \begin{align*} (3+i)^5&=(2e^{\frac{\pi}{6}i})^5\\ &=2^5e^{5\frac{\pi}{6}i}\\ &=32e^{\frac{5\pi}{6}i} \end{align*}$.

Finally, we convert this number back to the form $\displaystyle a+bi$.
$\displaystyle a=r\cos(\theta)=32\cos(\frac{5\pi}{6})=32(\frac{-\sqrt3}{2})=-16\sqrt3$
$\displaystyle b=r\sin(\theta)=32\sin(\frac{5\pi}{6})=32(\frac{1}{2})=16$

So our final answer is $\displaystyle -16\sqrt3+16i$.

Note that $\displaystyle \frac{\sqrt2}{2}+\frac{\sqrt2}{2}i$ lies in the first quadrant of the complex plane.

Any nonzero complex number $\displaystyle a+bi$ can be written in the form $\displaystyle re^{\theta i}$, where
$\displaystyle r^2=a^2+b^2$ and
$\displaystyle \tan(\theta)=\frac{b}{a}$.
(We stipulate that  $\displaystyle -\pi< \theta\leq\pi$ is in radians.)

Conversely, a nonzero complex number $\displaystyle re^{\theta i}$ can be written in the form $\displaystyle a+bi$, where
$\displaystyle a=r\cos(\theta)$ and
$\displaystyle b=r\sin(\theta)$.

We can convert $\displaystyle \frac{\sqrt2}{2}+\frac{\sqrt2}{2}i$ by using the formulas above:
$\displaystyle r^2=a^2+b^2=\left(\frac{\sqrt2}{2}\right)^2+\left(\frac{\sqrt2}{2}\right)^2=\frac{1}{2}+\frac{1}{2}=1$
$\displaystyle r=\sqrt1=1$,
and
$\displaystyle \tan(\theta)=\frac{b}{a}=\frac{\frac{\sqrt2}{2}}{\frac{\sqrt2}{2}}=1$
$\displaystyle \theta=\frac{-3\pi}{4},\frac{\pi}{4}$

Since $\displaystyle \frac{\pi}{4}$ lies in the first quadrant of the complex plane, as does $\displaystyle \frac{\sqrt2}{2}+\frac{\sqrt2}{2}i$, $\displaystyle \theta=\frac{\pi}{4}$.

So $\displaystyle \frac{\sqrt2}{2}+\frac{\sqrt2}{2}i=e^{\frac{\pi}{4}i}$.

We now substitute this into our original expression and expand.
$\displaystyle \begin{align*} \left(\frac{\sqrt2}{2}+\frac{\sqrt2}{2}\right)^{10}&=(e^{\frac{\pi}{4}i})^{10}\\ &=e^{\frac{10\pi}{4}i} \end{align*}$.
Because $\displaystyle -\pi< \theta\leq\pi$,  we substitute $\displaystyle \frac{10\pi}{4}$ with the coterminal angle $\displaystyle \frac{2\pi}{4}=\frac{\pi}{2}$.
$\displaystyle \left(\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i\right)^{10}=e^{\frac{\pi}{2}i}$.

Finally, we convert this number back to the form $\displaystyle a+bi$.
$\displaystyle a=r\cos(\theta)=\cos(\frac{\pi}{2})=0$
$\displaystyle b=r\sin(\theta)=\sin(\frac{\pi}{2})=1$

So our final answer is $\displaystyle 0+1i=i$.

Note that the $\displaystyle 1$ on the right side of the equation can be written as $\displaystyle e^{0i}$.

Multiplying the first two terms on the left side yields
$\displaystyle (-\sqrt2+\sqrt2i)(\frac{i}{4})=\frac{-\sqrt2}{4}-\frac{\sqrt2}{4}i$.
Note that this number lies in the third quadrant of the complex plane.

We now convert $\displaystyle -\frac{\sqrt2}{4}-\frac{\sqrt2}{4}i$ from the form $\displaystyle a+bi$ to the form $\displaystyle re^{i\theta}$using the identities
$\displaystyle r^2=a^2+b^2$ and
$\displaystyle \tan(\theta)=\frac{b}{a}$.
$\displaystyle r^2=a^2+b^2=\left(-\frac{\sqrt2}{4}\right)^2+\left(-\frac{\sqrt2}{4}\right)^2=\frac{2}{16}+\frac{2}{16}=\frac{4}{16}=\frac{1}{4}$
$\displaystyle r=\sqrt{\frac{1}{4}}=\frac{1}{2}$
$\displaystyle \tan(\theta)=\frac{b}{a}=\frac{-\frac{\sqrt2}{4}}{-\frac{\sqrt2}{4}}=1$
$\displaystyle \theta=-\frac{3\pi}{4},\frac{\pi}{4}$.
Since $\displaystyle -\frac{3\pi}{4}$ lies in the third quadrant of the complex plane, as does $\displaystyle -\frac{\sqrt2}{4}-\frac{\sqrt2}{4}i$, $\displaystyle \theta=-\frac{3\pi}{4}$.
So our new form is $\displaystyle \frac{1}{2}e^{-\frac{3\pi}{4}i}$.

Our equation now reduces to
$\displaystyle (\frac{1}{2}e^{-\frac{3\pi}{4}i})(z)=e^{0i}$.
We solve for z by dividing.
$\displaystyle \begin{align*} z&=\frac{e^{0i}}{\frac{1}{2}e^{-\frac{3\pi}{4}i}}\\ &=\frac{1}{\frac{1}{2}}e^{0i-\frac{-3\pi}{4}i}\\ &=2e^{\frac{3\pi}{4}i}. \end{align*}$

Finally, we convert this to the form $\displaystyle a+bi$ by using the identities
$\displaystyle a=r\cos(\theta)$ and
$\displaystyle b=r\sin(\theta)$.
$\displaystyle a=r\cos(\theta)=2\cos(\frac{3\pi}{4})=2(-\frac{\sqrt2}{2})=-\sqrt2$
$\displaystyle b=r\sin(\theta)=2\sin(\frac{3\pi}{4})=2(\frac{\sqrt2}{2})=\sqrt2$.

So our final answer is $\displaystyle z=-\sqrt2+\sqrt2i$.

$\displaystyle \overline{z}$ denotes the conjugate of $\displaystyle z$ and is defined as
$\displaystyle \ovelrine{z}=a-bi$.

Substituting this into the equation and simplifying yields:
$\displaystyle \begin{align*} a-bi&=-(a+bi)\\ a-bi&=-a-bi\\ a-bi+a+bi&=0\\ 2a&=0\\ a&=0 \end{align*}$

So the equation is only true if $\displaystyle a=0$.