Complex Functions & Solving Complex Equations Revisited

Basically,  complex functions  are analogous to regular (real)  functions, e.g.

f(x) = 3x – 4

f(x) = 2x² - 3x + 1

Except that complex variables of the form z = x + iy are incorporated. All the basic operations that apply to complex algebra, including polar forms, e.g.  r = exp[(i q)] ,  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) = e ^(-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) =  -iexp(-3x) sin (3y)

Finding numerical values for complex 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.

 

Find f(2i) for f(z) =  - 3z²

è       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 –  ½ i

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

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

3. Let f(z) = ln r + i(q)  where r = êz ê and q = Arg(z)

Find f(1):

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

Then:

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

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

Here, let z = r exp(i q)  then q = p/4

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

Therefore:

f (z) = exp(1)cos(p/4) +i(exp(1)sin(p/4) 

f(z) = exp[(cos(p/4) +i sin(p/4)] =  exp{Ö2/2 + iÖ2/2}

 f(z) =   1.922 + 1.922i

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) = z²  in the form f(z) = u(x,y) + iv(x,y)

If z = (x + iy) then:   z² = (x + iy)² = x² + i2xy  - y²  

=  (x² – y²) + 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) = x² + y²

Conversely, of course, one can be given the functions u(x,y) and v(x,y) then 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) =  = 4x² + i4y²

 Find f(z,z*)

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

Adding:

z + z* = (x + iy) + (x -  iy) = 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

It’s important to note that a polar form is also used, viz.

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

Example: Express f(z)= z²   in polar form.

z² = r² exp(i2(q)) = r² (cos (2q) + isin(2 q))

Therefore:

u(r, q) =  r²(cos (2q)    and:

v (r, q) =    r²(sin (2q)) 

These manipulations also allow one to see how to solve complex function equations..

For example: Solve for x and y in the equation:

3i +  6x   =   2  -   iy

As we've seen with the examples, one can work with the real part and the complex part separately.

Then:   6x = 2   so   x = 2 /6   =  1/ 3

And:    3i   =    -iy   or:    y  =   3i / (-i)  =   -3

Another example:

Solve for x and y:

4x² + i4y²    = 2 x   + yi

Then:     4x² =   2x   So:   x  =  2/ 4  =  1/2

And:  

i4y²    =   yi

So:    y   =   i /   4i     =   1/ 4   

Problems for the Math Maven:

1)  (a) Given u(x,y) + iv(x,y) = 2x² + i2y²  

find f(z,z*)

(b)  Express f(z)= z² + z – 3 in polar form

2) Solve for x and y:   8x 2 + 3iy  -  4    = 8y   – 4iy

3) Solve: (z+1) 3  = z 3