How can you use trigonometric functions to simplify $23 e^{\frac{2 π}{3} i}$ $23 e^( ( 2 pi)/3 i )$ into a non-exponential complex number?

Question

How can you use trigonometric functions to simplify $23 e^{\frac{2 π}{3} i}$ $23 e^( ( 2 pi)/3 i )$ into a non-exponential complex number?

1 Answer

Impact of this question

1365 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-14T07:32:47

-11.5 + i 11.5 $\sqrt{}$ $sqrt$ 3)

Explanation:

The polar form of a complex number (x, y) = x + i y is r $e^{i θ}$ $e ^(i theta)$ = r (cos( $θ$ $theta$ ) + i sin ( $θ$ $theta$ )). Here, r = 23 $and θ$ $and theta$ = 2 $π$ $pi$ /3. cos ( $2 π$ $2pi$ /3) = cos ( $π - \frac{π}{3}$ $pi - pi/3$ ) = - cos ( $\frac{π}{3}$ $pi/3$ ) and sin ( $2 π$ $2pi$ /3) = sins ( $π - \frac{π}{3}$ $pi - pi/3$ ) = sin ( $\frac{π}{3}$ $pi/3$ ).

Science

Math

Humanities

... and beyond

How can you use trigonometric functions to simplify $23 e^{\frac{2 π}{3} i}$ $23 e^( ( 2 pi)/3 i )$ into a non-exponential complex number?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How can you use trigonometric functions to simplify 23e2π3i 23 e^( ( 2 pi)/3 i ) into a non-exponential complex number?

1 Answer

Explanation:

Related questions

Impact of this question

How can you use trigonometric functions to simplify $23 e^{\frac{2 π}{3} i}$ $23 e^( ( 2 pi)/3 i )$ into a non-exponential complex number?