Lösung 3.1:4f

Because the equation contains both \displaystyle z and \displaystyle \bar{z}, we cannot use \displaystyle z (or \displaystyle \bar{z}) alone as an unknown, so we are forced to set

\displaystyle z=x+iy

and use the real part \displaystyle x and the imaginary part \displaystyle y as unknowns.

With this approach, the left-hand side of the equation becomes

\displaystyle (1+i)(x-iy)+i(x+iy) \begin{align} &=1\cdot x -1\cdot iy +i\cdot x -i^2y + i\cdot x + i^2y\\ &=x-iy+ix+y+ix-y\\ &=x+(2x-y)i\end{align}

and the whole equation becomes

\displaystyle x+(2x-y)i=3+5i.

The two complex numbers on the left- and right-hand sides are equal when both real and imaginary parts are equal, i.e.

\displaystyle \begin{cases}x=3\\2x-y=5\end{cases}.

This gives \displaystyle x=3 and \displaystyle y=2x-5=2\cdot 3-5=1. Thus, the equation has the solution \displaystyle z=3+i.

A quick check shows that \displaystyle z=3+i satisfies the equation in the exercise:

\displaystyle \begin{align}LHS &= (1+i)\bar{z}+iz=(1+i)(3-i)+i(3+i)\\ &= 3-i+3i+1+3i-1 = 3+5i = RHS\end{align}