Lösung 3.1:4f

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

z=x+iy

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

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

(1+i)(x−iy)+i(x+iy)=1x−1iy+ix−i2y+ix+i2y=x−iy+ix+y+ix−y=x+(2x−y)i

and the whole equation becomes

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.

x=32x−y=5 

This gives x=3 and y=2x−5=23−5=1. Thus, the equation has the solution z=3+i.

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

LHS=(1+i)z+iz=(1+i)(3−i)+i(3+i)=3−i+3i+1+3i−1=3+5i=RHS