Question-189786

Question Number 189786 by sonukgindia last updated on 21/Mar/23

Answered by a.lgnaoui last updated on 22/Mar/23

A ∙ z^{z} = {(e^{i θ})}^{e^{i θ}} = {(\cos θ + i \sin θ)}^{z}

= e^{z lnz} = e^{e^{i θ} \ln (e^{i θ})} = e^{\ln [{(e^{i θ})}^{(\cos θ + i \sin θ)}]}

= e^{[\ln (e^{i θ \cos θ} \times e^{- θ \sin θ}]} = e^{[\ln (e^{i θ \cos θ} \times \ln (e^{- θ \sin θ)}]}

= \frac{\ln (e^{i θ \cos θ})}{\ln (e^{θ \sin θ})} = \frac{\cos (θ \cos θ) + i \sin (θ \cos θ)}{θ \sin θ}

= \frac{\cos (θ \cos θ)}{θ \sin θ} + i \frac{\sin (θ \cos θ)}{θ \sin θ}

\Rightarrow

z^{z} = e^{[\frac{\cos (θ \cos θ}{θ \sin θ}} \times e^{i \frac{\sin (θ \cos θ}{θ \sin θ}}

= e^{\frac{\cos (θ \cos θ)}{θ \sin θ}} [\cos (\frac{\sin (θ \cos θ)}{θ \sin θ}) + \sin (\frac{\sin (θ \cos θ}{θ \sin θ})]

\Rightarrow

R (z^{z}) = e^{\frac{\cos (θ \cos θ)}{θ \sin θ}} \times \cos^{(\frac{\sin (θ \cos θ)}{θ \sin θ})}