If z=reiθ , then |eiz|=

Question

If z=reiθ , then |eiz|= (A) 1 (B) e2rsinθ (C) ersinθ (D) e-rsinθ

Tardigrade Admin · Accepted Answer

z=reiθ =r(cos θ+i sin θ) ∴ iz=ir(cos θ+i sin θ)=ir cos θ+i2 r sin θ =-r sin θ+ir cos θ |eiz|=|e(-r sin θ +ir cos θ)|=|e-r sin θ|⋅|eir cos θ| =e-r sin θ [√cos2(r cos θ)+sin2(r cos θ)] =e-r sin θ ×1=e-r sin θ

Q. If $z = r e^{i θ}$ , then $∣ ∣ e^{i z} ∣ ∣ =$

$1$

$e^{2 rs in θ}$

$e^{rs in θ}$

$e^{- rs in θ}$

Q. If z=reiθ , then ∣∣​eiz∣∣​=

1

e2rsinθ

ersinθ

e−rsinθ

z=reiθ=r(cosθ+isinθ) ∴iz=ir(cosθ+isinθ)=ircosθ+i2rsinθ =−rsinθ+ircosθ ∣∣​eiz∣∣​=∣∣​e(−rsinθ+ircosθ)∣∣​=∣∣​e−rsinθ∣∣​⋅∣∣​eircosθ∣∣​ =e−rsinθ[cos2(rcosθ)+sin2(rcosθ)​] =e−rsinθ×1=e−rsinθ

Q. If $z = r e^{i θ}$ , then $∣ ∣ e^{i z} ∣ ∣ =$

$1$

$e^{2 rs in θ}$

$e^{rs in θ}$

$e^{- rs in θ}$