Imaginary integrals => Real ones

9 May, 2020

Okay it's 6 steps and if you do it enough times it wil be down to only 5 steps.

Just so all of this actually makes sense, we will be using the following example:

f(z) = \frac{1}{z}

Above equation is to be integrated along the circle|z|=R, starting and finishing on the route.

Before getting into the sateps lets make use we know the basics

Basic Definition

An imaginary number is conventionally denoted as

z = x + \iota y

and a function is denoted as

f(z) = u(x, y) + \iota v(x, y)

here u and v are functions of x and y, and that i thing is iota also known as root of -1.

1) Path

Write the path in terms of only one variable, conventionally t. So, here we can write 'z' as:

z = R \cos t + \iota R \sin t

from this we can directly compare and get x and y

x = R \cos t

y = R \sin t

That was all step 1 was.

2) Function to x y

Now we will get function which is in terms of 'u' and 'v' into terms of 'x' and 'y'.

f(z) = \frac{1}{z}
f(z) = \frac{1}{x + \iota y}
f(z) = \frac{x-\iota y}{x^2 + y^2}

From this we can take out values of 'u' and 'v' as

u = \frac{x}{x^2+y^2};v = -\frac{y}{x^2+y^2}

3) uv to t

At this step substitute the earlier values of 'u' and 'v' we have in terms of 'x' and 'y', to have it in terms to 't'.

u = \frac{\cos t} {R}
v = -\frac{\sin t} {R}
We are almost done now.

4) Piecing it all together

Now plug them into this equation you can do the following.

\int{f(z)} = \int{u\frac{dx}{dt}dt} - \int{v\frac{dy}{dt}dt} + \iota\int{u\frac{dy}{dt}dt} +\iota\int{u\frac{dx}{dt}dt}

5) Precaution

In the very likely case you forget the equation:

You can get it directly from this:

\int{f(z)dz} = \int{(u+\iota v)(dx + \iota dy)}

Just multiply those two brackets in the L.H.S and you'll be at the equation above it in 2 lines.