Multiplication of Complex Numbers
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Key Questions

First, complex numbers can come in a variety of forms!
Ex: multiply
#3i*4i =# Remember, with multiplication you can rearrange the order (called the Commutative Property):
#3*4*i*i =12i^2# ... and then always substitute 1 for
#i^2# :#12*1 = 12# Ex: the numbers might come in a radical form:
#sqrt(3)*4sqrt(12) =# You should always "factor" out the imaginary part from the square roots like this:
#sqrt(1)sqrt(3)*4*sqrt(1)sqrt(4)sqrt(3) =# and simplify again:
#=i*4*sqrt(3)*sqrt(3)*sqrt(4)#
#=i*4*3*2 = 24i# Ex: what about the Distributive Property?
#3i(4i  6) =# #=12i^2 18i#
#=12(1)  18i#
#= 12  18i# And last but not least, a pair of binomials in a + bi form:
Ex: (3  2i)(4 + i) =
=12 + 3i  8i 
#2i^2#
= 12  2(1) + 3i  8i
= 12 + 2  5i
= 14  5i 
Let
#z_1# and#z_2# be two complex numbers.By rewriting in exponential form,
#{(z_1=r_1e^{i theta_1}),(z_2=r_2 e^{i theta_2}):}# So,
#z_1 cdot z_2 =r_1e^{i theta_1}cdot r_2 e^{i theta_2} =(r_1 cdot r_2)e^{i(theta_1+theta_2)}# Hence, the product of two complex numbers can be geometrically interpreted as the combination of the product of their absolute values (
#r_1 cdot r_2# ) and the sum of their angles (#theta_1+theta_2# ) as shown below.
I hope that this was clear.

In trigonometric form, a complex number looks like this:
#a + bi = c*cis(theta)#
where#a# ,#b# and#c# are scalars.Let two complex numbers:
#> k_(1) = c_(1)*cis(alpha)#
#> k_(2) = c_(2)*cis(beta)#
#k_(1)*k_(2) = c_(1) * c_(2) * cis(alpha) * cis(beta) =#
#= c_(1) * c_(2) * (cos(alpha) + i*sin(alpha)) * (cos(beta) + i*sin(beta))# This product will end up leading to the expression
#k_(1)*k_(2) =#
#= c_(1)*c_(2)*(cos(alpha + beta) + i*sin(alpha + beta)) =#
#= c_(1)*c_(2)*cis(alpha+beta)# By analyzing the steps above, we can infer that, for having used generic terms
#c_(1)# ,#c_(2)# ,#alpha# and#beta# , the formula of the product of two complex numbers in trigonometric form is:
#(c_(1) * cis(alpha)) * (c_(2) * cis(beta)) = c_(1)*c_(2)*cis(alpha+beta)# Hope it helps.

To explain this, I will name two generic complex.
#c_1 = a*cis(alpha)# and#c_2 = b*cis(beta)# The product between
#c_1# and#c_2# is:
#ab*cis(alpha)cis(beta) =#
#ab*(cos(alpha)+isin(alpha)) (cos(beta)+isin(beta)) =#
#ab*({cos(alpha)cos(beta)sin(alpha)sin(beta)} +#
#{i(sin(alpha)sin(beta)+cos(alpha)sin(beta)}) =#
#ab*{cos(a+b)+isin(a+b)}# //Therefore, we can assume that the product of the two complex numbers
#c_1# and#c_2# can be generaly given by the form above.Ex.:
#(2*cis(pi)) * (3*cis(2pi)) = 6*cis(3pi) = 6*cis(pi)# Hope it helps.

Complex numbers are multiplied simply by the rules of classical algebra .
let,
#Z_1# = a+ib and#Z_2# =x+iy be two complex numbers.lets Z be another complex number such that,
Z=
#Z_1# #*# #Z_2# #:.# Z= (a+ib)#*# (x+iy)
#rArr# Z= ax+ iay+ ibx+#i^2# by.
#rArr# Z= ax + i(ay+bx) by....................... [as i=#sqrt# 1,#:.# #i^2# =1]
#rArr# Z= (axby) + i(ay+by) , which is another complex number.