\(\displaystyle \overline{z}\) 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 \(\displaystyle a=0\).

As messy as it looks, this is just a geometric series.

we will use the partial sum formula for the geometric series.

\(\displaystyle S_{n}=\frac{a_{1}(1-r^n)}{1-r}\)

\(\displaystyle a_{1}=e^{2\pi i/m}\)

\(\displaystyle r=e^{2\pi i/m}\)

\(\displaystyle S_{m}=\frac{a_{1}(1-r^m)}{1-r}=\frac{e^{2\pi i/m}(1-{\color{Red} (e^{2\pi i/m})^m})}{1-e^{2\pi i/m}}\)

the red part is the only part that matters....the \(\displaystyle m's\) cancel out leaving....

\(\displaystyle S_{m}=\frac{a_{1}(1-r^m)}{1-r}=\frac{e^{2\pi i/m}(1-{\color{Red} (e^{2\pi i})})}{1-e^{2\pi i/m}}\)

and...

\(\displaystyle e^{2\pi i}=1\)

thus we have...

\(\displaystyle S_{m}=\frac{a_{1}(1-r^m)}{1-r}=\frac{e^{2\pi i/m}(1-{\color{Red} (1)})}{1-e^{2\pi i/m}}\)

which gives the answer of zero.

\(\displaystyle S_{m}=0\)

Applying the definition of derivative, we have that

\(\displaystyle f'(z) = \lim_{\Delta z \to 0} \frac{f(z+\Delta z) - f(z)}{\Delta z} \\\)

\(\displaystyle f'(z) = \lim_{\Delta z \to 0}\frac{|z+\Delta z|^2 - |z|^2}{\Delta z} = \lim_{\Delta z \to 0} \frac{(z+\Delta z)(\bar{z}+\overline{\Delta z}) - z\overline{z}}{\Delta z} = \lim_{\Delta z \to 0} \overline{z} + \overline{\Delta z} + z \frac{\overline{\Delta z}}{\Delta z}\)

If the limits exists, it can be found by letting \(\displaystyle \Delta z\) approach \(\displaystyle 0\) in any manner. 

In particular, if we it approach through the points \(\displaystyle (\Delta x , 0 )\), we have that 

\(\displaystyle \overline{\Delta z} = \overline{\Delta x + i0} = \Delta x - i0 = \Delta x + i0 = \Delta z\)

A similar approach with \(\displaystyle (0, \Delta y)\) implies that \(\displaystyle \overline{\Delta z } = -\Delta z\)

Since limits are unique, these two approaches imply that 

\(\displaystyle \overline{z} + z = \overline{z} - z\), which implies \(\displaystyle z=0\) and \(\displaystyle f'(z)\) cannot exist when \(\displaystyle z \neq 0\)

\(\displaystyle v(x,y)\)is said to be a harmonic conjugate of \(\displaystyle u(x,y)\) if their are both harmonic in their domain and their first order partial derivatives satisfy the Cauchy-Riemann Equations. Computing the partial derivatives

\(\displaystyle u_x(x,y) = 2-3x^2+3y^2 = v_y(x,y)\)

\(\displaystyle v(x,y) = 2y-3x^2y+y^3 + C\)

where \(\displaystyle C\) is any arbitrary constant.

Rewriting \(\displaystyle f(z)\) in real and complex components, we have that

\(\displaystyle f(z) = z- \bar{z} = (x+yi) - (x-yi) = 2yi\)

So this implies that \(\displaystyle f(z) = f(x + yi) = u(x,y) + v(x,y)i\)

where \(\displaystyle u(x,y) = 0 \text{ and } v(x,y) = 2y\)

Therefore, checking the Cauchy-Riemann Equations, we have that

\(\displaystyle u_x(x,y) = 0 \neq v_y(x,y) = 2\)

So the Cauchy-Riemann equations are never satisfied on the entire complex plane, so \(\displaystyle f(z)\) is differentiable nowhere.