The value of e 2- i isA- e 2-cos 1- i sin 1-B- cos 1- i sin 1C- e -2-cos 1- i sin 1-D- e -cos 1-i sin 1-

Name: JEE- Grade 11- Mathematics- Complex Numbers- Session 04- D01
Uploaded: 2022-07-04T10:36:42+05:30
Description: We have, e^iθ= cosθ+isinθ ⇒ e^2+i=e^2· e^i =e^2(cos 1 + i sin 1) ...

We have, e^iθ= cosθ+isinθ ⇒ e^2+i=e^2· e^i =e^2(cos 1 + i sin 1) ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Euler's Representation

The value of ...

Question

The value of $e^{2 + i}$ is

e [cos 1 + i sin 1]

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

e^{2} [cos 1 + i sin 1]

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

cos 1 + i sin 1

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

e^{- 2} [cos 1 + i sin 1]

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

Open in App

Solution

The correct option is B $e^{2} [cos 1 + i sin 1]$
We have, $e^{i θ} = cos θ + i sin θ$
$\Rightarrow e^{2 + i} = e^{2} \cdot e^{i} = e^{2} (cos 1 + i sin 1)$

Suggest Corrections

Similar questions

Q. If

2 \sqrt{2} x^{4} = (\sqrt{3} - 1) + i (\sqrt{3} + 1),

then

x = cos \frac{1}{4} (2 n π + k) + i sin \frac{1}{4} (2 n π + k); n = 0, 1, 2, 3,

where

k =

Q. The biggest among

(sin 1 + cos 1), (\sqrt{sin 1} + \sqrt{cos 1}), (sin 1 - cos 1)

lim n \to \infty n^{2} ⎛ ⎜ ⎜ ⎝ \sqrt{(1 - c o s \frac{1}{n}) \sqrt{(1 - c o s \frac{1}{n}) \sqrt{(1 - c o s \frac{1}{n}) . . . . . \infty}}} ⎞ ⎟ ⎟ ⎠

is equalte

Q. If

A, B, C, D

are the angle of a quadrilateral, prove that

cos \frac{1}{2} (A + B) + cos \frac{1}{2} (C + D) = 0

Q. The greatest among (sin1+cos1),

(\sqrt{s i n 1} + \sqrt{c o s 1}), (s i n 1 - c o s - 1)

and 1 is

Join BYJU'S Learning Program

Explore more

Euler's Representation

Standard XII Mathematics

Join BYJU'S Learning Program

The value of e2+i is

The correct option is B e2[cos1+isin1]We have, eiθ=cosθ+isinθ ⇒e2+i=e2⋅ei=e2(cos1+isin1)

The value of $e^{2 + i}$ is

The correct option is B $e^{2} [cos 1 + i sin 1]$
We have, $e^{i θ} = cos θ + i sin θ$
$\Rightarrow e^{2 + i} = e^{2} \cdot e^{i} = e^{2} (cos 1 + i sin 1)$