How do you multiply #e^(( 11 pi )/ 8 i) * e^( pi/2 i ) # in trigonometric form?

1 Answer
Jan 3, 2016

Euler formula #e^(itheta) = cos(theta)+isin(theta)# that would convert to trigonometric form. When we multiply trigonometric form we add the angles and multiply the modulus.

Explanation:

#e^((11pi)/8i)*e^((pi/2)i)#

#e^((11pi)/8i) = cos((11pi)/8) + isin((11pi)/8)#

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

#e^((11pi)/8i)*e^((pi/2)i)#

# = (cos((11pi)/8) + isin((11pi)/8))*(cos(pi/2) + isin(pi/2))#

#=cos((11pi)/8+pi/2)+isin((11pi)/8+pi/2)#

#=cos((11pi)/8+(4pi)/8)+isin((11pi)/8+(4pi)/4)#

#=cos((11+4)pi/8) + isin((11+4)pi/8)#

#=cos((15pi)/8)+isin((15pi)/8)#

Note: The question said in trigonometric form so converted to trigonometric form and multiplied. Using Euler's form would be easy multiply and then convert, the choice is yours.

Alternate method:
#e^((11pi)/8i)*e^((pi/2)i)#
#=e^(((11pi)/8+pi/2)i)# using exponent rule #a^m*a^n=a^(m+n)#
#=e^(((15pi)/8)i)#

Use the Euler's formula

#=cos((15pi)/8) + isin((15pi)/8)# Answer.