## Monday, March 4, 2013

### Solutions to Polar Forms for Complex Numbers (2)

1)We need to use the form r exp(iq) to express the following:

a) (2 + 3i)(1 – 2i)

Solution:

Multiplying the two factors we get:

2 + 3i -6(i2) -4(i) = 8 + 3i – 4i = 8 –i

First, convert to standard polar form, given :

z = x + iy = 8 - i

so: r = r = [82 + (-1)2]1/2 = Ö65  =   8.1

q = arg(z) = arctan (y/x) = arctan (-1/8) = -7.1 deg.

So: z = 8.1 [cos(-7.1) + isin(-7.1)]

Or: z = 8.1 exp (i(-7.1))

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

Expand to get: 1 + 1i - 1i -(i)(i)  = 1 - (-1) =  1 + 1 = 2

For which arg(z) = 0  (No imaginary component, i)

And arctan (0) = 0 and there is no polar form.

Okay, well let me refine that! One can obtain r = Ö2.

Then z =  r exp (i 0) = Ö2 (1) = Ö2

c) (1 + Ö-3)2

Expand to get::

z = (1 + Ö-3) (1 + Ö-3)
But bear in mind that: (Ö-3) = [iÖ3)] so we have:

(1 + [iÖ3]) x (1 + [i Ö3]) = 1 + 2i[Ö3)] + (-1)3 = -2 + i2[iÖ3]

z = -2 + 3.46i

r = [(-2)2 + (2(iÖ3) 2] =  [4 + 36]1/2 = [40]1/2 = 6.3

q = arctan (y/x) = arctan (2[Ö3)]/ -2) = arctan (-Ö3) = (-60) = -p/3
Thus, z = 6.3 exp(i(-p/3))

2) Plot the results of (b) and (c) on the same Argand Diagram.

Solution: Since for 1(b)  arctan (0) = 0  then q = 0 so there is only one complex  number to be identified on the Argand diagram: -2 + 3.46i. The other value is 2. So we plot '2' on the real (x-) axis and then (-2, 3.46) with -2 on the (- x) axis and 3.46 on the +y- axis.

This yields a parallelogram with points at: (0, 0), (2, 0), (-2, 3.46) and (0, 3.46) . Sketching the resultant - one will draw it from (0,0) to (0, 3.46). It will be seen to have a length of 3.46 units.

Check using algebra: The resultant of 2 + (-2 + 3.46i) = 3.46i   or conforming to what the sketch for the Argand diagram gives: a vector from (0,0) directly up the y-axis and 3.46 units in length.