A) \[i{{e}^{-{{e}^{-iz}}}}\]
B) \[i{{e}^{-{{e}^{iz}}}}\]
C) \[-i{{e}^{-{{e}^{-iz}}}}\]
D) None of these
Correct Answer: B
Solution :
[b] \[(f-ig)(z)=f(z)-ig(z)=sinz-icosz\] \[=-i(cosz+isinz)=-i{{e}^{iz}}=\theta (say)\] Now \[{{(f+ig)}^{*}}(f-ig)(z)=(f+ig)(f-ig)(z)\] \[=(f+ig)(\theta )=f(\theta )+ig(\theta )=sin\theta +icos\theta \] \[=i(cos\theta -isin\theta )=i{{e}^{-i\theta }}=i{{e}^{-i(-i{{e}^{iz}})}}=i{{e}^{-{{e}^{iz}}}}\]
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