If z - re i - then - e iz - is equal toA- e-r cos-B- e-C- e - r sin-D- e-i sin-

Z = re^i θ = r(cosθ + isinθ) ⇒ iz = ir (cosθ + isinθ) ⇒ iz= rsinθ + ircosθ ⇒ e^iz= e^ r sinθ + ircosθ = e^ rsinθ e^ricosθ ⇒ |e^iz| = |e^ rsinθ||e^r icosθ| = e ...

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Standard XII

Mathematics

Modulus of a Complex Number

If z = re i θ...

Question

If $z = r e^{i θ}$ , then $| e^{i z} |$ is equal to

e^{- r sin θ}

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e^{- i sin θ}

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e^{- r cos θ}

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e^{- θ}

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Solution

The correct option is A $e^{- r sin θ}$
$z = r e^{i θ} = r (cos θ + i sin θ)$
$\Rightarrow i z = i r (cos θ + i sin θ)$
$\Rightarrow i z = - r sin θ + i r cos θ$
$\Rightarrow e^{i z} = e^{- r sin θ + i r cos θ}$
$= e^{- r sin θ} e^{r i cos θ}$
$\Rightarrow | e^{i z} | = | e^{- r sin θ} | | e^{r i cos θ} |$
$= e^{- r sin θ} | e^{i α} |$ , where $α = r cos θ$
$= e^{- r sin θ} [{cos}^{2} α + {sin}^{2} α]^{\frac{1}{2}}$
$= e^{- r sin θ}$

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Modulus of a Complex Number

Standard XII Mathematics

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If z=reiθ, then |eiz| is equal to

The correct option is A e−rsinθz=reiθ=r(cosθ+isinθ) ⇒iz=ir(cosθ+isinθ) ⇒iz=−rsinθ+ircosθ ⇒eiz=e−rsinθ+ircosθ =e−rsinθericosθ ⇒|eiz|=|e−rsinθ||ericosθ| =e−rsinθ|eiα|, where α=rcosθ =e−rsinθ[cos2α+sin2α]12 =e−rsinθ

If $z = r e^{i θ}$ , then $| e^{i z} |$ is equal to