All the magic numbers in one place!

Algebra Level 2

e 1 − i π = ? \large{e^{1-i\pi}=?}

i i 1 − i 1-i 1 + i 1+i − i -i e e − e -e

Md Omur Faruque by MD Omur Faruque , 25, Bangladesh

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Md Omur Faruque
Oct 10, 2015

From Euler's Theorem, we know that, e i θ = cos ⁡ θ + i sin ⁡ θ e^{i\theta}= \cos \theta+i\sin \theta

So, e 1 − i π = e × e i ( − π ) e^{1-i\pi}= \color{#3D99F6}{e}\times e^{i(-\pi)} = e × [ cos ⁡ ( − π ) + i sin ⁡ ( − π ) ] = \color{#3D99F6}{e}\times [ \cos (-\pi)+i\sin (-\pi)] = e × ( − 1 + i × 0 ) = \color{#3D99F6}{e}\times(-1+i\times0) = e × ( − 1 ) = − e = \color{#3D99F6}{e}\times (-1)=\color{#0C6AC7}{\boxed{-e}}

9 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice work. This formula helps in solving such type of problems quickly and with accuracy

Akhil Krishna - 5 years, 8 months ago
Maadhav Gupta
Oct 8, 2017

Knowledge of Euler's Identity is sufficient here. (That is, e i π = − 1 e^{i \pi}=-1 )

e 1 − i π = e 1 × e − i π = e e i π = e − 1 = − e e^{1- i \pi} = e^1 \times e^{-i \pi } = \frac {e}{e^{i \pi }} = \frac {e}{-1} = -e

1 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...