**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.