Linear algebra : Solving $i \cdot\bar{z} = 2 +2i$ $i\cdot\bar{z} = 2+2i$
I know that $\bar{z} = a-bi$ so then i get $i(a-bi)=2+2i$
Then $ai+b=2+2i$ (because $i^2=-1$)
When 2 complex numbers are equal you usually can equal their parts
Ex: $2+2i=a+bi$ so $a=2$ and $b=2$
But in this case I aint got $a+bi$ I have $ai+b$
The answer is $z=2-2i$ but i dont understand how to get there
Thanks
 A: It looks like you are trying to solve
$$
i\bar{z} = 2 + 2i.
$$
You can start by multiplying both sides by $i$:
$$
i^2\bar{z} = 2i + 2i^2 = -2 + 2i
$$
This gives you
$$
-\bar{z} = -2 + 2i.
$$
or
$$
\bar{z} = 2 - 2i
$$
Now just take the complex conjugate on both sides:
$$
z = \overline{\bar{z}} = \overline{2 - 2i} = 2 + 2i.
$$
A: It might be useful also to think of the problem geometrically. The operation $z \mapsto i\overline{z}$ takes the vector $z$, reflects it across the $x$-axis (this is $z \mapsto \overline{z}$) and then rotates it by $90$ degrees counterclockwise (this is $w \mapsto iw$). Solving $i\overline{z} = 2 + 2i$ amounts to finding a complex number $z$ such that if $z$ is reflected and then rotated, we get $2 + 2i$. Since $2 + 2i$ lies in the first quadrant and makes an angle of $45$ degrees with the $x$-axis, after we reflect it and rotate it, we return to the same complex number and so $z = 2 + 2i$.
A: Suppose $z=a+bi$; then $\bar{z}=a-bi$ and $i\bar{z}=ai+b=b+ai$. Thus the real part of $i\bar{z}$ is $b$ and the imaginary part is $a$.
Hence, equating them gives $b=2$ and $a=2$.
However, you can do it more easily: since $i^{-1}=-i$, multiplying both sides by $-i$ gives
$$
\bar{z}=-i(2+2i)=2-2i
$$
and so
$$
z=\overline{2-2i}=2+2i
$$
A: We have: $i\cdot\bar{z}=2+2i$
Let $z=a+bi \Rightarrow \bar{z}=a-bi$:
$\Rightarrow i\cdot(a-bi)=2+2i$
$\Rightarrow ai-bi^{2}=2+2i$
$\Rightarrow ai+b=2+2i$
$\Rightarrow b+ai=2+2i$
Comparing real and imaginary parts:
$\therefore\hspace{5 mm}b=2$ and $a=2$
