Question
If $z=r e^{i \theta}$ and $w=e^{i z}$ then $(\ln |w|)^{2}+(\arg w)^{2}$(a) 0(b) 2(c) $2|z|^{2}$(d) $|z|^{2}$
Step 1
Step 1: Given that $z=r e^{i \theta}$ and $w=e^{i z}$, we can substitute the value of $z$ in $w$ to get $w=e^{i r e^{i \theta}}$. Show more…
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