How do you multiply #e^(( 2 pi )/ 3 i) * e^( 3 pi/2 i ) # in trigonometric form?

1 Answer
Yonas Yohannes
Apr 7, 2016

#C_12=e^[((2 pi)/3 +(3pi)/2)] = e^ [(13pi)/6i]#

Explanation:

Given: Two complex numbers,
#C_1=e^((2 pi)/ 3 i), C_2= e^((3 pi)/2i)#

Required: The product, #C_1*C_2=e^((2 pi)/ 3 i)*e^((3 pi)/2i)#

Solution Strategy: Use complex multiplication mechanism using the general exponential product rule, #e^(ai)* e^(bi) = e^(ai+bi)#
Thus;
#C_(12)=C_1*C_2=e^((2 pi)/ 3 i)*e^((3 pi)/2i)=e^[((2 pi)/ 3 i)+((3pi)/2i)]#
#C_12=e^[((2 pi)/3 +(3pi)/2)] = e^ [(13pi)/6i]#

Related questions
See all questions in The Trigonometric Form of Complex Numbers
Impact of this question
1524 views around the world
You can reuse this answer
Creative Commons License