If $z = 1 + i$ , then the argument of $z^{2} e^{z - i}$ is

\frac{π}{2}

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

\frac{π}{6}

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

\frac{π}{4}

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

\frac{π}{3}

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

0

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

Open in App

Solution

The correct option is D $\frac{π}{4}$
Let $w = z^{2} e^{z - i}$
Put $z = 1 + i$
$\Rightarrow w = {(1 + i)}^{2} e^{1 + i - i} = (1 + 2 i + i^{2}) e$
$= (1 + 2 i - 1) e = 2 e i$
Now, argument of $w = a r g (w)$
$= {tan}^{- 1} (\frac{2 e}{0}) = {tan}^{- 1} \infty = \frac{π}{2}$

Suggest Corrections

Similar questions

z,

| z - 1 + i | + | z + i | = 1,

z

(

\in (- π, π])

k + | k + z^{2} | = | z |^{2}, (k \in R^{-})

z

z = \frac{- 2 (1 + 2 i)}{3 + i}

i = \sqrt{- 1}

θ (- π < θ \leq π)

z

z = \frac{(1 + i) (1 + 2 i) (1 + 3 i) \dots \dots \dots (1 + n i)}{(1 - i) (2 - i) (3 - i) \dots \dots \dots (n - i)}

i = \sqrt{- 1}, n \in N

z

If z be a non- real complex number satisfying |z|=2, then which of the following is/are true?

Join BYJU'S Learning Program