Lösung 3.1:4d

In the equation, z occurs only as z and, to begin with, we can therefore treat z as unknown.

Divide both sides by 2+i,

z=2+i1+i

and calculate the quotient on the right-hand side by multiplying top and bottom by the complex conjugate of the numerator:

z=(2+i)(2−i)(1+i)(2−i)=22−i212−1i+i2−ii=4+12−i+2i+1=53+i=53+51i

This means that z=53−51i

We check that z=53−51i satisfies the original equation:

LHS=(2+i)z=(2+i)(53−51i)=(2+i)(53+51i)=253+251i+i53+i51i=56+52i+53i−51=56−1+52+3i=1+i=RHS