How can you use trigonometric functions to simplify $8 e^( ( 7 pi)/4 i )$ into a non-exponential complex number?

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Answer 1 · 2015-12-28T18:51:02

$8*e^((7pi)/4*i)=4pi-4pi*i$

If we have a complex number in trigonometric form:

$z=|z|*e^(i*varphi)$ ,

then according to de Moivre formula we can write, that:

$|z|*e^(i*varphi)=|z|*(cosvarphi+i*sinvarphi)$

If we assume that the module $|z|$ is equal to $1$ , then the formula simplifies to:

$e^(i*varphi)=(cosvarphi+i*sinvarphi)$

So the expression $8*e^((7pi)/4*i)$ can be written as:

$8*(cos((7pi)/4)+i*sin((7pi)/4))=$

$8*[cos(2pi-pi/4)+i*sin(2pi-pi/4)]=$

$8*[cos(pi/4)-i*sin(pi/4)]=8*[sqrt(2)/2-i*sqrt(2)/2]=$

$4sqrt(2)-4sqrt(2)*i$

How can you use trigonometric functions to simplify 8 e^( ( 7 pi)/4 i ) 8 e^( ( 7 pi)/4 i ) into a non-exponential complex number?