How do you multiply #e^(( 9 pi )/ 8 i) * e^( pi/2 i ) # in trigonometric form?

1 Answer
Alan P.
Dec 30, 2016

#e^((9pi)/8i) * e^(pi/2i)=color(green)(sin(pi/8)-icos(pi/8))#

Explanation:

Using Euler's formula
#color(white)("XXX")e^((9pi)/8i) = cos((9pi)/8)+i * sin((9pi)/8)#

#color(white)("XXXXX")=cos(pi+pi/8)+i * sin(pi+pi/8)#

#color(white)("XXXXX")=-cos(pi/8)+i * (-sin(pi/8))#

#color(white)("XXXXX")=-cos(pi/8)- i * sin(pi/8)#

and
#color(white)("XXX")e^(pi/2i)=cos(pi/2)+ i * sin(pi/2)#

#color(white)("XXXXX")=0 + i * 1#

#color(white)("XXXXX")=i#

Therefore
#color(white)("XXX")e^((9pi)/8i) * e^(pi/2i) = [-cos(pi/8)- i * sin(pi/8)] * i#

#color(white)("XXXXXXX")=- i * cos(pi/8) - (-1) * sin(pi/8)#

#color(white)("XXXXXXX")=sin(pi/8)-icos(pi/8)#

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