How do you evaluate $e^{\frac{3 π}{2} i} - e^{\frac{π}{2} i}$ $e^( (3 pi)/2 i) - e^( ( pi)/2 i)$ using trigonometric functions?

Question

How do you evaluate $e^{\frac{3 π}{2} i} - e^{\frac{π}{2} i}$ $e^( (3 pi)/2 i) - e^( ( pi)/2 i)$ using trigonometric functions?

1 Answer

Impact of this question

4749 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-04T07:20:58

$- 2 i$ $-2i$

Explanation:

$e^{i x} = cos x + i sin x$ $e^(ix)=cos x + i sin x$ .
$cos (3 \frac{π}{2}) = cos (\frac{π}{2}) = 0$ $cos(3pi/2)=cos(pi/2)=0$
$sin (3 \frac{π}{2}) = sin (π + \frac{π}{2}) = - sin (\frac{π}{2}) = - 1$ $sin(3pi/2)=sin(pi+pi/2)=-sin(pi/2)=-1$ and $sin (\frac{π}{2}) = 1$ $sin(pi/2)=1$
So the expression simplifies to $- i - i = - 2 i$ $-i-i=-2i$ .(

Science

Math

Humanities

... and beyond

How do you evaluate $e^{\frac{3 π}{2} i} - e^{\frac{π}{2} i}$ $e^( (3 pi)/2 i) - e^( ( pi)/2 i)$ using trigonometric functions?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you evaluate e3π2i−eπ2i e^( (3 pi)/2 i) - e^( ( pi)/2 i) using trigonometric functions?

1 Answer

Explanation:

Related questions

Impact of this question

How do you evaluate $e^{\frac{3 π}{2} i} - e^{\frac{π}{2} i}$ $e^( (3 pi)/2 i) - e^( ( pi)/2 i)$ using trigonometric functions?