Solving $z+i\overline{z}=iz-\overline{z}$ I want to solve $z+i\overline{z}=iz-\overline{z}$ ($\overline{z}$ is the complex conjugate). I have solved it setting $z=a+bi$. But can one solve without writing it $z$ a certain form, factorization maybe?
Thanks in advance
 A: Let's rewrite the equation:
$$
z+i\bar z=iz-\bar z \Longleftrightarrow \bar z(i+1)=z(i-1)\Longleftrightarrow z=\frac{i+1}{i-1}\bar z=-i\bar z\;\;\;.
$$
Writing now $z=re^{i\theta}$, and observing that $i=e^{i\frac{\pi}2}$ the last equation becomes
$$
re^{i\theta}=e^{i\frac{\pi}2}re^{-i\theta}\;\;\;
$$
i.e.
$$
e^{i\theta}=e^{i(\frac{\pi}2-\theta)}\;\;.
$$
So the solutions are $z=re^{i\theta}$, for $r\ge0$ and $\theta\in\Bbb R$ such that $\theta=\frac{\pi}2-\theta+2k\pi$, i.e. $\theta=\frac{\pi}4+k\pi$.
So you have infinite solutions in modulus, but your argument must be $\frac{\pi}4$ or $\frac{5\pi}4$.
Thus the solutions are $z_r=r(1+i)$, $r\in\Bbb R$.
A: Hint For $w = z + i \bar{z}$ we have
$$\bar{w} = \overline{z + i\bar{z}} = \bar{z} - i z = -(iz - \bar{z}),$$ in which case we can rewrite the given equation as
$$w = -\bar{w},$$
which is satisfied iff $w$ is purely imaginary.
A: Alternatively (the benefit of hindsight helps here), one can define $\zeta$ by setting $$z = \zeta e^{-\pi i / 4},$$
in which case substituting in the original equation and rearranging gives
$$2 \sqrt{2} \Im \zeta = 0.$$
This is satisfied iff $\zeta$ is real, so the solution set is
$$\{t e^{-\pi i / 4} : t \in \Bbb R\}.$$
This is the line produced by rotating the real line about the origin $\frac{\pi}{4}$ radians clockwise, namely, the line through the origin that bisects the second and fourth quadrants.
A: You could write it $z(1-i)=i\bar z(i-1)$...
EDIT: since there is a question about where does it come from...
It is coming from: $z+i\bar z=iz -\bar z$ that is $z-iz=-i\bar z -\bar z$, then $z(1-i)=-i(\bar z-i\bar z)=-i\bar z(1-i)$
A: From your given condition it can be inferred that, 
$$\frac{z+\bar z}{z-\bar z}=i$$ 
Using componendo and dividendo,
$$\frac{z}{\bar z}=\frac{1+i}{i-1}$$
$$z=-i\bar z$$
A: Another answer that uses an important and fundamental property:
Note, $$z+\overline{z} = 2\operatorname{Re}z,$$ and $$z-\overline{z} = 2i\operatorname{Im}z.$$
Using this, we get
$$z+i\overline{z} = iz-\overline{z} \implies z+\overline{z} = i(z-\overline{z}).$$
This becomes
$$2\operatorname{Re}z = i(2i\operatorname{Im}z)\\
\operatorname{Re}z = -\operatorname{Im}z.$$
Therefore, solutions must lie along the line $y=-x$ in the complex plane.
Let's check using rectangular form. $z = a+bi$.
$$a+bi+i(a-bi) = i(a+bi)-a+bi.$$
Let $b = -a$.
$$a-ai+i(a+ai) = i(a-ai)-a-ai \\
a-ai+ai-a = ai+a-a-ai \\
0 = 0.$$
A: $$z+i\overline{z}=iz-\overline{z}\Longrightarrow$$
$$(a+bi)+(a-bi)i=(a+bi)i-(a-bi)\Longleftrightarrow$$
$$(a+b)(1+i)=(a+b)(-1+i)\Longleftrightarrow$$
$$a+b+ai+bi=-a-b+ai+bi\Longleftrightarrow$$
$$2a+2b=0\Longleftrightarrow$$
$$2(a+b)=0\Longleftrightarrow$$
$$a+b=0\Longleftrightarrow$$
$$\{_{b=-a}^{a=-b}$$
