1.

If `z=r e^(itheta)`, then prove that `|e^(i z)|=e^(-r s inthetadot)`

Answer» `z = re^(i0) = r (cos theta + i sin theta)`
`rArr iz = ir (cos theta + i sin theta)`
`=- rsin theta + ir cos theta`
`rArr e^(1z) = e^((-rsin theta+ ir cos theta))`
`= e^((-rsin theta)) e^((ri cos theta))`
`rArr |e^(iz)|=|e^(-rsin theta)||e^(r icos theta)|`
`= e^((- r sin theta))|e^(ialpha)|," ""where"alpha = r cos theta`
`= e^(-rsin theta)`


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