Multiplication of Complex Numbers:
How do we multiply two complex numbers?
Assume that we want to multiply two complex numbers:
#color(green)((a+bi)*(c+di))#
We can use the FOIL method to multiply:
Firsts: #a * c#
Outers: #a * di#
Inners: #bi * c#
Lasts: #bi * di#
Hence,
#color(green)((a+bi)*(c+di)=ac+adi+bci+bdi^2)#
Now, we will consider our problem and multiply the complex numbers given to us:
#color(blue)((1-2i)(-2-3i)#
#rArr [1 * (-2)] + [1 (-3i)] + [(-2i) (-2)] + [(-2i) * (-3i)]#
#rArr (-2) + (-3i) + (+4i) + (+6i^2)#
#rArr (-2) + (+1i) + (6i^2)#
Note that #color(red)(" "i^2 = -1)#
Hence,
#rArr (-2) + (+1i) + [6*(-1)]#
#rArr (-2) + (+1i) + (-6)#
#rArr (-8 +i)#
Hence, our intermediate answer:
#color(blue)((1-2i)(-2-3i)=-8 +i#
We also know that, in trigonometric form of complex numbers
#Z = x + iy#
#r = |Z| = sqrt(x ^ 2 + y ^2#
#x = r Cos (Theta)#
#y = r Sin (Theta)#
#Z = r[Cos(Theta) + i Sin(Theta)]#
We will now calculatre
#r = |Z| = sqrt(x ^ 2 + y ^2#
#rArr |Z| = sqrt(1^2 + (-8)^2)#
#rArr |Z| = sqrt(65)#
#Theta = arctan(-1/8)#
#Theta = -3.2659#
We have, #Z = r[Cos(Theta) + i Sin(Theta)]#
#Z = sqrt(65)[Cos(-3.2659) + i Sin(-3.2659)]#
Hope this helps.
