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:
- 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
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