Lösung 3.3:4d

Aus Online Mathematik Brückenkurs 2

To avoid having z in the denominator, we multiply both sides of the equation by z:

1+z2=21z

This multiplication could possibly introduce a false root if it turns out that the new equation has z=0 as a root. The old equation, for understandable reasons, have z=0 as a solution.

If we move the terms over to the left-hand side and complete the square, we get

z2−21z+1=0z−412−412+1=0z−412+1615=0

This gives that the equation has solutions

z=41+i415  and z=41−i415 

None of these solutions are equal to zero, so these are also solutions to the original equation.

We substitute the solutions into the original equations to assure ourselves that we have calculated correctly.

z=41−i415:LHS=z1+z==141−i1415+41−i415=41+i41541−i41541+i415+41−i415=116+161541+i415+41−i415=41+i415+41−i415=21=RHS

z=41+i415:LHS=z1+z==141+i1415+41+i415=41−i41541+i41541−i415+41+i415=116+161541−i415+41+i415=41−i415+41+i415=21=RHS