How do you multiply #e^(( 5 pi )/ 4 i) * e^( pi/2 i ) # in trigonometric form?

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Answer 1 · 2016-04-13T12:35:50

#e^((5pi/4)i)*e^((pi/2)i)=-1/sqrt2-i1/sqrt2#

A complex number can be written in polar form in two ways - either as #r*e^(itheta)# or as #rcostheta+irsintheta#.

Hence,

#e^((5pi/4)i)=cos(5pi/4)+isin(5pi/4)# and

#e^((pi/2)i)=cos(pi/2)+isin(pi/2)#

Hence, #e^((5pi/4)i)*e^((pi/2)i)#

= #(cos(5pi/4)+isin(5pi/4))*(cos(pi/2)+isin(pi/2))#

= #(cos(5pi/4)+isin(5pi/4))*(0+i)#

as #cos(pi/2)=0# and #sin(pi/2)=1#

= #(icos(5pi/4)+i^2sin(5pi/4))#

= #-sin(5pi/4)+icos(5pi/4)#

= #-1/sqrt2-i1/sqrt2#