If z=r $e^{i θ}$ ,then | $e^{i z}$ |=

$e^{r s i n θ}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$e^{- r s i n θ}$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

$e^{- r c o s θ}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$e^{r c o s θ}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B
$e^{- r s i n θ}$

if z=r $e^{i θ}$ =r(cos $θ$ + isin $θ$ )

$\Rightarrow$ iz=ir(cos $θ$ + isin $θ$ )=-rcos $θ$ + irsin $θ$

or $e^{i z}$ = $e^{- r c o s θ + i r s i n θ}$ = $e^{- s i n θ} e^{r i c o s θ}$

or | $e^{i z}$ |= $| e^{- r s i n θ} | | e^{i r c o s θ} |$

= $e^{- r s i n θ}$ $[c o s^{2} (r c o s θ) + s i n^{2} (r c o s θ)]^{2}$ = $e^{- r s i n θ}$

Suggest Corrections

Similar questions

If z=r $e^{i θ}$ ,then | $e^{i z}$ |=

z = r e^{i θ}

| e^{i Z} | =

z = r e^{i θ}

| e^{i z} |

z = r e^{i θ}

| e^{i z} |

z = r e^{i θ}

∣ ∣ e^{i z} ∣ ∣ = e^{- r} s i n θ

Join BYJU'S Learning Program