Back
in March we examined a few examples of complex functions. Basically, it
entailed writing them in assorted different forms, including complex polar. We
operated more or less at the cusp between beginner complex analysis and more
meaty stuff – peculiar to intermediate analysis. Now, in this post, I want to
look at the interface between basic and more advanced complex analysis as
embodied in the Cauchy – Riemann equations.

Basically,
full details are beyond the scope of one or even a few blog posts, but
interested readers are invited to get any number of texts which treat the
topic, including ‘

*Complex Analysis for Mathematics and Engineering*’, ‘*Complex Variables and Applications’*, and ‘*Applied Complex Variables’*(Harper Collins College outline).
Basically,
given a function defined: f(z)
= f(x + iy)

= u(x,y) + iv(x,y)

= u(x,y) + iv(x,y)

which
is taken to be differentiable at the point z

_{0}= x_{0 }+ iy_{0}- then the partial derivatives of u and v exist at the point (x_{0 }, y_{0}) and satisfy the equations:
a)
¶ u/ ¶ x = ¶ v/ ¶ y and
b) ¶ v/ ¶ x =
- ¶ u/ ¶ y

Which
are the Cauchy- Riemann equations. If such condition is met then the function
is said to be analytic in the region, Â.

An
additional condition likely to be examined is whether the function is harmonic.
If it is, then u and v also have continuous 2

^{nd}partial derivatives. Then, we can differentiate both sides of (a) with respect to x, and (b) with respect to y to obtain:
1)
¶

^{2}u/ ¶ x^{2}= ¶^{2}v/ ¶ x ¶ y and
2) - ¶

^{2}u/ ¶ y^{2}= ¶^{2}v/ ¶ y ¶ x
From
which:

¶

^{2}u/ ¶ x^{2}= - ¶^{2}u/ ¶ y^{2}or ¶^{2}u/ ¶ x^{2}+ ¶^{2}u/ ¶ y^{2}= 0
Which
is known as

**.**Laplace ’s equation
Similarly,
we can perform an analogous process for v (differentiating both sides of (a)
with respect to y and (b) with respect to x) to arrive at:

¶

^{2}v/ ¶ x^{2}+ ¶^{2}v/ ¶ y^{2}= 0
If
then this equation is satisfied, v is harmonic.

Let’s
look at an example, using a function from the March 19 blog.

We
had: f(z) = (x

^{2 }– y^{2}) + i2xy
So that: u(x,y) = x

^{2}- y^{2}
And
v(x,y) = 2x y

Now,
we first check to see if the eqns. are analytic

Take
¶ u/ ¶ x =
2x

And: ¶ v/ ¶ y
= 2x

Since: ¶ u/ ¶ x
= ¶ v/ ¶ y then u(x.y) is analytic

Now
check the other function, v(x.y)

¶ v/ ¶ x = 2y

And: - ¶ u/ ¶ y = -
(-2y) = 2y

So
that v(x,y) is analytic..

Note:
If f(z) is analytic everywhere in the complex plane it is said to be

*an entire function*.
Now,
check to see whether the functions are

*harmonic*.
For
u(x,y) this means we need: ¶

^{2}u/ ¶ x^{2}+ ¶^{2}u/ ¶ y^{2}= 0
Since:
¶ u/ ¶ x =
2x, then ¶

^{2}u/ ¶ x^{2}= 2
Since:
¶ u/ ¶ y =
-2y then ¶

^{2}u/ ¶ y^{2}= -2
Then:

¶

^{2}u/ ¶ x^{2}+ ¶^{2}u/ ¶ y^{2}= 2 + (-2) = 0
For
v(x.y): we need ¶

^{2}v/ ¶ x^{2}+ ¶^{2}v/ ¶ y^{2}= 0
Since:
¶ v/ ¶ x = 2y
then ¶

^{2}v/ ¶ x^{2}= 2
Since:
¶ v/ ¶ y =
2x then ¶

^{2}v/ ¶ y^{2}= 2
Then:
¶

^{2}v/ ¶ x^{2}+ ¶^{2}v/ ¶ y^{2}= 2 + 2 = 4
So,
v(x,y) is evidently not harmonic since the sum of the 2

^{nd}partials are not equal zero.
Problems
for the Math Maven:

1)
Given
the function: u(x,y) = x

^{3}– 3xy^{2}
Show
the function is harmonic on the entire complex plane.

2)
Given
the function: u(x.y) = exp(-x) [x sin y – y cos y]

a)
Show
u(x,y) is harmonic

b)
Find
v(x,y) such that f(z) = u + iv is analytic

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