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))(a+bi)⋅(c+di)
We can use the FOIL method to multiply:
Firsts: a * ca⋅c
Outers: a * dia⋅di
Inners: bi * cbi⋅c
Lasts: bi * dibi⋅di
Hence,
color(green)((a+bi)*(c+di)=ac+adi+bci+bdi^2)(a+bi)⋅(c+di)=ac+adi+bci+bdi2
Now, we will consider our problem and multiply the complex numbers given to us:
color(blue)((1-2i)(-2-3i)(1−2i)(−2−3i)
rArr [1 * (-2)] + [1 (-3i)] + [(-2i) (-2)] + [(-2i) * (-3i)]⇒[1⋅(−2)]+[1(−3i)]+[(−2i)(−2)]+[(−2i)⋅(−3i)]
rArr (-2) + (-3i) + (+4i) + (+6i^2)⇒(−2)+(−3i)+(+4i)+(+6i2)
rArr (-2) + (+1i) + (6i^2)⇒(−2)+(+1i)+(6i2)
Note that color(red)(" "i^2 = -1) i2=−1
Hence,
rArr (-2) + (+1i) + [6*(-1)]⇒(−2)+(+1i)+[6⋅(−1)]
rArr (-2) + (+1i) + (-6)⇒(−2)+(+1i)+(−6)
rArr (-8 +i)⇒(−8+i)
Hence, our intermediate answer:
color(blue)((1-2i)(-2-3i)=-8 +i(1−2i)(−2−3i)=−8+i
We also know that, in trigonometric form of complex numbers
Z = x + iyZ=x+iy
r = |Z| = sqrt(x ^ 2 + y ^2r=|Z|=√x2+y2
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.
