## Thursday, February 28, 2013

### Another Form for Complex Numbers (1)

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:

x + iy = r(cos (Θ) + isin(Θ))

And 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)

Now we look 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

B = z2 = -2 -3i

So: z1 = x1 + iy1

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

So (Θ1) = -45 degrees = -π /4

Now 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:  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)

Now, how do we 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))

[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

Problems for the Math Maven:

1) Based on the resultant for (x3, y3) obtain arg(z3)  and thence the polar form for the resultant. (Hint: Remember  arg(z3)  =  Θ )

2) Now combine this resultant z3 with that from C (Fig. 1) which we call z4, to obtain z5

Thence, find arg(z5) and write in polar form.

3) Obtain the complex product for [z3• z4]