Saturday, April 29, 2017

Math Revisited: Complex Analysis (1) In advanced physics, including plasma physics and quantum mechanics, the importance of complex numbers and analysis can't be over emphasized. Consider the diagram shown with three complex numbers identified.   To extend the generality of complex numbers and enhance their applicability, it's useful to write them in what's called "polar form".

A critical part is finding the angle shown, referred to as the argument. We can see from the diagram 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))

To get r:

r = [x2 + y2]1/2 =  [42 + 32]1/2 = 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

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 = 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

Complex Division:
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:

(
q1q2)   = 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    (q1q2).

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

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)
r1/r2 = Ö2/ 2

Then subtract angles: [(
q1q2) ] = {(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

Another very convenient way to express complex numbers is in the exponential form.

Thus, we can write: cos(
q) + isin(q) = r exp (iq )

Thus, the previous numbers we divided (z1 and z2) may be expressed:

z1 =
Ö2 [cos (45) + isin(45)] = Ö2 exp (i p/4)

z2 = 2[cos(-30) +isin(-30)] = 2 exp(i (-
p/6))

Practice Problems:

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

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