How can you use trigonometric functions to simplify $7 e^( ( 3 pi)/4 i )$ into a non-exponential complex number?

Question

1361 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-20T17:47:39

The answer is $=-(7sqrt2)/2+(i7sqrt2)/2$

We know that $e^(itheta)=costheta+isintheta$

So, $7e^(3pii/4)=7(cos ((3pi)/4)+isin((3pi)/4))$

$cos(3pi/4)=-sqrt2/2$

$sin(3pi/4)=sqrt2/2$

Therefore,

$7e^(3pii/4)=7(-sqrt2/2+isqrt2/2)$

$-(7sqrt2)/2+(i7sqrt2)/2$

How can you use trigonometric functions to simplify 7 e^( ( 3 pi)/4 i ) 7 e^( ( 3 pi)/4 i ) into a non-exponential complex number?