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Lösung 3.1:4c

Aus Online Mathematik Brückenkurs 2

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Version vom 10:32, 11. Mär. 2009

If we subtract 2z from both sides,

iz+22z=3

and then subtract 2 from both sides, we have z terms left only on the left-hand side,

iz2z=32.

After taking out a factor z from the left-hand side,

(i2)z=5

we obtain, after dividing by 2+i,

z=52+i=5(2i)(2+i)(2i)=(2)2i2(5)(2)5(i)=4+110+5i=510+5i=2+i.

A quick check shows that z=2+i satisfies the original equation,

LHSRHS=iz+2=i(2+i)+2=2i1+2=1+2i=2z3=2(2+i)3=4+2i3=1+2i.