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#