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If z=x+iy and w=(1-iz)/(z-i), then mod w=1 implies ,that in the complex plane

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Solution:

\left | omega 
ight |=1 Rightarrow left | (1-iz)/(z-i) 
ight |\ \Rightarrow left | -i^2-iz 
ight |=left | z-i 
ight |\ \Rightarrow left | (-i)(z+i) 
ight |=left | z-i 
ight |\ \Rightarrow left | z+i 
ight |=left | z-i 
ight |

z lies on the perpendicular bisector of the segment  joining i and -i

z lies on the real axis.

Posted by

Deependra Verma

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