Complex Functions (II)

We’ll begin by solving the problems 1,3, 5 from the earlier assignment in complex functions:

1) Find f(1+i) for f(z) = 1/(z^2 + 1)

Then: f(z) = 1/ {(1 +i)^2 + 1} = 1 /(2i+1) = 0.2 - 0.4i


2) f(-2) for f(z) = ln r + i(theta)

where r = abs(z) and theta = Arg(z)

Then: r = abs(-2) = 2 and Arg z = Arg(-2)

So theta = 116.57 deg

So: f(-2) = ln (2) + i(116.57 deg)


3) solve: (z+1)^3 = z^3

Expand the left side and set equal to the right:

z^3 + 3z^2 +3z + 1 = z^3

so:

3z^2 + 3z = -1 or 3z^2 + 3z +1 =0

(which can be solved using the quadratic formula, to give two roots)

z1 = ½ + i(3)^1/2/ 6 and z2 = -(½ ) + i(3)^1/2/ 6

checking the result against the equation:

z^3 = 0.192i and (z^3 + 1)^3 = 0.192i


Now let’s continue to look at more aspects of complex functions. Specifically, functions of a complex variable can also be written in the form:

f(x + iy) = u + iv

and since u,v depend on x and y, they can be considered as real functions of the real variables x and y such that:

u = u(x,y) and v = v(x,y)

Example: write f(z) = z^2 in the form f(z) = u(x,y) + iv(x,y)

If z = (x + iy) then z^2 = (x + iy)^2 = x^2 + i2xy - y^2 = (x^2 – y^2) + i2xy

The last step above shows how the complex function is separated into two parts, one with the factor i, the other without. The one with the factor applies to the function v(x,y) so:

v(x,y) = 2xy

While: u(x,y) = x^2 + y^2


Conversely, of course, one can be given the functions u(x,y) and v(x,y) and be asked to find f(z), e.g. in terms of z and-or its complex conjugate, z*.

Example: Given u(x,y) + iv(x,y) = 4x^2 + i4y^2 find f(z,z*)

Again, let: z = x + iy, and z* = x – iy

Adding:

z + z* = 2x

so we see: x = (z + z*)/2

Now, subtracting: (z – z*) = [x + iy – x + iy] = i2y

So: y = (z – z*)/ 2i

Since we have both x and y we can now formulate the function f(z,z*):

f(z,z*) = 4[(z + z*)/2]^2 + i4[(z – z*)/ 2i]^2


Now, before leaving the basics of complex functions, it’s important to note that a polar form is also used, viz.

f (z) = f(r exp(i(theta)) = u(r,theta) + iv(r,theta)

Example: Express f(z)= z^2 + z – 3 in polar form

z^2 = r^2 exp(i2(theta)) = r^2 (cos (2theta) + isin(2 theta))

z = r exp(i(theta)) = r(cos(theta) + isin(theta)

so: z^2 + z = r^2 (cos (2theta) + isin(2 theta)) + r(cos(theta) + isin(theta)

collecting like terms in i and simplifying:


f(z) = r^2(cos (2theta) + r(cos(theta)) + i{sin(2theta) + sin(theta)} – 3

so: iv(r,theta) = i{sin(2theta) + sin(theta)}

and v (r,theta) = {sin(2theta) + sin(theta)}

while:

u(r,theta) = r^2(cos (2theta) + r(cos(theta)) – 3


Problems:

1) Express f(z) = z*^2 + (2-3i)z in (u +iv) format
2) Express: z^5 + z*^5 in polar format.