Complex Numbers - Find z such that $(z+1)^4=(\bar{z}+i)^4$ I have the following problem and I don't know how to resolve it:

$(z+1)^4=(\bar{z}+i)^4$

Tried to expand the equation and work with $z=a+b*i$ and $\bar{z}=a-b*i$ but I came to a dead end
Any suggestions will be much appreciated.
Thanks
 A: You can take (all) the fourth roots of both sides to show $z$ must be a solution to one (or more) of the following four equations:
\begin{align*}
z + 1 &= \overline{z} + i \\
z + 1 &= i(\overline{z} + i) \\
z + 1 &= -(\overline{z} + i) \\
z + 1 &= -i(\overline{z} + i) \\
\end{align*}
That might make things easier!
A: First put $z=x+yi$, where $x,y \in \mathbb{R}$. Ok, so you have $$\left(\frac{z+1}{\overline{z}+i}\right)^4=1.$$ So $\frac{z+1}{\overline{z}+i}$ has 4 possible values (which are the 4 roots of the unit, namely $1, -1, e^{\frac{\pi}{2}i}=i, e^{\frac{3\pi}{2}i}=-i$.) Just plug-in each value and solve for $x$ and $y$.
A: Hint:  $z=i$ is obviously not a solution, so it is safe to divide by $(\bar z+i)^4\ne 0\,$. With $w=\cfrac{z+1}{\bar z + i}\,$ the equation then becomes $w^4=1\iff (w^2-1)(w^2+1)=0$ with solutions $w=\pm 1, \pm i$.
To reverse the substitution, one needs to solve $z+1=w(\overline z +i)$ for $z$, which implies:
$$\require{cancel}
z=w(\overline z +i)-1=w\big(\,\overline{w(\bar z + i)-1}+i\,\big) - 1= w\big(\overline w(z-i)-1+i\big) -1 \\[3px] \iff\quad \bcancel{z} = \bcancel{z} - i -w + w i-1 \quad\iff\quad w=-i
$$
Therefore the only eligible solution is $w=-i$, then the equation becomes $z+\bcancel{1}=-i(\overline z +\bcancel{i})\,$ $\iff z+i\bar z =0\,$ which has the second bisector $\operatorname{Re}(z)+\operatorname{Im}(z)=0$ as the solution set.
A: $(z+1)^4 = (\overline{z} + i)^4 \implies |z+1|^4 = |\overline{z} + i|^4 \implies |z+1|^2 = |\overline{z} + i|^2$.
If we let $z = x + iy$, then the above equation reduces to $y = -x$. Thus, the solution must lie on the line $x = t, y = -t, \; \forall t \in \mathbb{R}$.
Now we need to find $t$ such that $z = t -it$ satisfies the original equation.
Using above, we get $z + 1 = (1 + t) -it \;$ and $\; \overline{z} + i = t + i(1+t)$. Notice that $i(z+1) = \overline{z} + i$. Take fourth powers and use $i^4 = 1$. Thus the equation is satisfied by all $t \in \mathbb{R}$.
Thus solutions are all the points on the line $x+y=0$.
A: The equation $$(z+1)^4 = (\overline{z}+1)^4$$ is equivalent to $$\left( \frac{z+1}{\overline{z}+1} \right)^4 =1.$$ Since $w= \left (\frac{z+1}{\overline{z}+1} \right)^4 \in \mathbb{C}$ then you need to find $w^4=1$ i.e. the four 4th-roots of the unity; which are $\pm 1$ and $\pm i$. Put $z=a+bi$ with $a$ and $b$ real number and solve the four equations.
