Introducing Complex Functions (1)

Having spent some five blogs on basic complex algebra we’re now in a position to examine complex functions. Basically these are like regular function, e.g.

f(x) = 3x – 4

f(x) = 2x^2 + 2x -1

except that complex variables of the form z = x + iy are incorporated. All the basic operations we saw that apply to complex algebra, including polar forms (in sin(theta), cos(theta) or = exp[(i theta)] ) apply also to complex functions. Note that since complex functions are dependent on the complex variable z, we typically write them as f(z).

A good way to get started is by applying simple operations to functions.

For example, let f(z) = exp(-3z)

Find the real and imaginary parts of the function f(z)

Since z = x + iy, we may write:

f(z) = exp[-3(x + iy)] = exp(-3x) [exp(-i3y)]

and

f(z) = exp(-3x)[cos (3y) – isin(3y)]

Then : Re f(z) = Re exp(-3z) = exp(-3x)cos (3y)

And : Im f(z) = Im exp(-3z) = -exp(-3x) sin (3y)

Finding numerical values for functions in many ways resembles the way we do it for real functions, simply substitute the f-value to be found into the function f(z), viz.

f(2) for f(x) = 2(exp(x)) = 2 exp(2) = 2(7.389) = 14.796

f(1/2) for f(x) = ln(x) = ln(1/2) = - 0.693

f(3) for f(x) = 1/x^3 = 1/(3)^3 = 1/27

Now, some complex function values:

E,g, Find f(2i) for f(z) = - 3z^2

f(2i) = -3(2i)^2 = -3 (-4) = 12

More examples:

1. Find: f(-3i) for f(z) = (z + 2 – 3i)/ (z + 4 – i)

Again: f(-3i) = {(-3i +2 -3i)/ (-3i + 4 – i)}

= (2 – 6i)/ (4 – 4i) = 1 – i ½

2. Find f(2i -3) for f(z) = (z + 3)^2(z – 5i)^2

so: f(2i -3) = {(2i -3)+3}^2 (2i – 3 – 5i)^2 = {(-4)(18i)} = -72i

3. Let f(z) = ln r + i(theta) where r = abs(z) and theta = Arg(z)

Find f(1)

f(1) = ln 1 + i Arg(1) but we know that theta = pi/4 for Arg (1)

e.g. arctan(1) = 45 deg = pi/4

Then:

f(1) = ln 1 + i(pi/4)

4. f(i pi/4) for f(z) = exp(x) cos(y) + i(exp(x)sin(y)

Here, let z = r exp(i theta) then theta = pi/4

And exp(i pi/4)= cos(pi/4) + isin(pi/4) with r = 1

Therefore:

f (z) = exp(1)cos(pi/4) +i(exp(1)sin(pi/4) = i2.718(2 (2)^1/2/2) = 3.844i


Problems:

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

2. Repeat for f(z) = z + z^-2 +5

3. Find: f(-2) for f(z) = ln r + i(theta), where r = abs(z) and theta = Arg(z)

4. Find f(1 +ipi/4) for f(z) = exp(x)[cos(y) + isin(y)]

5. Solve: (z + 1)^3 = z^3