Dealing With The Basics Of Complex Numbers Pt.1 - The Polar Form

 Consider the diagram shown below with three complex numbers identified:

                        Complex numbers shown on Cartesian graph.

To extend the generality of complex numbers and enhance their applicability, it's useful to write them in what's called "polar form".  This instalment and the next will deal with that treatment. A critical part is finding the angle shown, referred to as the argument. We can see from the diagram (Fig. 1) that the angle Θ may be found using:

arctan (y/x) = arctan(3/4) = 36.8 deg

Thus, Θ = 36.8 degrees is the argument

Now, any complex number (x + iy) may be written in polar form such that:

  x + iy = r(cos (Θ) + isin(Θ))

To get r:   r = [x² + y²]^1/2 =  [4² + 3²]^1/2 = [25]^1/2 = 5

Therefore we may write:  

 (x + iy) = 5(cos (36.8) + isin(36.8))

Note there is also the abbreviated function (based on the combo of sine and cosine):    cis (Θ) = cos (Θ) + isin(Θ)

So we can finally write:  C = r cis(Θ) = 5 cis (36.8) 

We look now at the vectors A and B, which we’ll henceforth call z1 and z2 to be consistent with complex notation. Our eventual goal will be to find the resultant, which will come in the next installment. In the meantime we will be working toward showing the multiplication and division of two complex forms, call them: z1 and z2, e.g. [z1 + z2].

From the diagram:

A= z1 = -2 + 2i,   and:   B = z2 = -2 -3i

So: z1 = x1 + iy1

Thence:   arg(z1) = arctan(y1/x1) = arctan (-2/2) = arctan(-1)

So (Θ1) = -45 degrees = -π /4

Next  find r1:   r₁ =[x₁² + y₁²]^1/2 =  [1 + 1]^1/2 = Ö2   

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

We now turn to the vector B which is expressed:  z2 = x2 + iy2= -2 -3i

then: arg(z2) = arctan(y2/x2) = arctan (-3/-2) = arctan (3/2) = 56.3 deg

While:   r₂ = [x₂² + y₂²]^1/2 =  [(-2)² + (-3)²]^1/2 = [13]^1/2 =  3.6
Therefore:

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

Next we seek to obtain the complex product: [z1•z2]

We have that:  [z1•z2] = (z1•z2) cis(arg(z1) – arg(z2))

But:   (z1•z2) = Ö2 (3.6) = 5.1

And:   arg(z1) – arg(z2) = (-45) – (56.3) = -101.3

So that:
[z1•z2] = 5.1 cis(-101.3) = 5.1 (cos (-101.3) + isin(-101.3))

Simplifying:   [z1•z2] = 5.1((-0.195) + i(-0,98))

[z1•z2] = 0.99 + 0.98i

To get the resultant: z1 + z2 = z3:

A + B = z1 + z2 =[ (-2 + 2i) + (-2 – 3i)] = -4 –i

In any case: x3 + iy3 = - 4 – i

Suggested Problems:

1) Express each of the following end results in the form:       r exp(iq):

a) (2 + 3i)(1 – 2i)

b) (1 + i) (1- i)

c) (1 + Ö-3)²

2) Plot the results of (b) and (c) on the same Argand diagram and obtain the resultant. Check algebraically!