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OpenStudy (anonymous):
if |z+1|=z+2(1+i),find z?
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OpenStudy (helder_edwin):
put \(z=a+ib\). do all the operations on each side and you will get a system of equations
OpenStudy (anonymous):
When you distribute the 2 to the (1+i) you get |z+1|=z+2+2i. Since 2i=-1, the equation becomes |z+1|=z+2-1 which would simplify to |z+1|=z+1
OpenStudy (primeralph):
Nope.
OpenStudy (primeralph):
2i is not -1.
OpenStudy (anonymous):
oh crap. youre right. sorry...
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OpenStudy (primeralph):
|dw:1399769884103:dw|
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