## Sunday, March 17, 2013

### Introducing Complex Functions (1)

Having spent several blogs on basic complex algebra and roots, we’re now in a position to examine complex functions. Basically these 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 we saw that apply to complex algebra, including polar forms, including  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 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

Problems for the Math Maven:

1. Find f(1+i) for: f(z) = 1/ (z2 + 1)

2.  Find: f(-2) for f(z) =  ln r + i(q), where r = êz ê  and q = Arg(z)

3.   Solve: (z + 1)3 = z3