Dealing With The Basics Of Complex Numbers (Part 2): Division In Polar Form

 Let's say we want to divide:

z1 = Ö2(cos(-45) + isin(-45)) = Ö2 cis(-45)

by

z2 = 3.6(cos(56.3) + isin(56.3)) = 3.6 cis(56.3)

In all such cases of complex division we require that the z, r in the denominator 

not be zero.

Thus:

(z₁/z₂) =  (r₁ cis(q₁)/ r₂ cis(q₂)) = (r₁/ r₂) cis (q₁ –  q₂)

Now: (r1/ r2) = (1.414/ 3.6) = 0.39

And we saw previously:

(q₁ – q₂)   = arg(z₁) – arg(z₂) = (-45) – (56.3) = -101.3

Thus, the basic procedure for division entails dividing the lengths (r’s) and subtracting the angles (q₁ – q₂).

So:

(z₁/ z₂) =  0.39 (cos (-101.3) + isin(-101.3))

= 0.39((-0.195) + i(-0.98)) = -0.07 + 0.38i

What about?   (1 + i)  ¸  Ö3  – i

The first order of business is to get dividend and divisor each into polar form, specifically as a (cis) function:

Then (1 + i) = z₁  = x1 + iy1, so arg(z₁ ) = arctan (y1/x1)

Further:
arctan (y1/x1) = arctan (1/1) = arctan (1) so q₁ = 45 deg

What about r1?    r1= [1² + 1²]^1/2 =  Ö2 = 1.4

so z₁ = 1.4 [cos (45) + isin(45)] = 1.4 cis(45)

Now: z₂ = Ö3  – i    So:

 arg(z₂) = arctan(y₂/x₂) =  arctan(-1/ Ö3) so q₂ = (-30 deg)

And for r₂: r₂ = [(Ö3)² + (-1)²]^1/2  = Ö4 = 2

Then: z₂ = 2[cos(-30) +isin(-30)] = 2cis(-30)
We divide:  (z₁/z₂)

Which means dividing the r’s first:

r₁/r₂ = Ö2/ 2 

Then subtract angles: [(q₁ – q₂) ] = {(45 deg) – (-30 deg)} = 75 degrees

So the end result of the division is:

(z₁/z₂)  =  Ö2/ 2   cis(75) = Ö2/ 2  {cos(75) + isin(75)}

= 0.707{cos(75) + isin(75)}

Since cos(75) = 0.258 and sin(75) =0.966, we have:

(z₁/z₂)   = 0.707[(0.258) + i(0.966)] = 0.183 + 0.683i

Problem:

Use the polar form of complex numbers to divide:

z1 = -2 + 2i by z2 = -2 - 3i