If z=r e^i θ , then |e^i z |=

Correct answer: (D) e^-r sin θ. We have, z =r e^i θ=r(cos θ+i sin θ) ∴ i z =i r(cos θ+i sin θ) =i r cos θ+i^2 r sin θ=-r sin θ+i r cos θ |e^i z | = |e^(-r sin θ+i r cos θ) | = |e^-r sin θ | ·|i r cos θ| =e^-r sin θ [√cos^2(r cos θ)+sin^2(r cos θ) ] =e^-r sin θ × 1=e^-r sin θ

Q72 / 180 · COMEDK 2017 · Mathematics

If $z=r e^{i \theta}$, then $\left|e^{i z}\right|=$

A
1
B
$e^{2 r \sin \theta}$
C
$e^{r \sin \theta}$
Correct answer
D
$e^{-r \sin \theta}$

Solution

We have, $\begin{aligned} z &=r e^{i \theta}=r(\cos \theta+i \sin \theta) \\ \therefore i z &=i r(\cos \theta+i \sin \theta) \\ &=i r \cos \theta+i^{2} r \sin \theta=-r \sin \theta+i r \cos \theta \\ \left|e^{i z}\right| &=\left|e^{(-r \sin \theta+i r \cos \theta)}\right| \\ &=\left|e^{-r \sin \theta}\right| \cdot|i r \cos \theta| \\ &=e^{-r \sin \theta}\left[\sqrt{\cos ^{2}(r \cos \theta)+\sin ^{2}(r \cos \theta)}\right] \\ &=e^{-r \sin \theta} \times 1=e^{-r \sin \theta} \end{aligned}$

Hence, the correct answer is option: 'D'.

If $z=r e^{i \theta}$, then $\left|e^{i z}\right|=$

Solution

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