# Abhishek's thoughts • Converting Imaginary Integrals to real ones

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# Converting Imaginary Integrals to real ones

Okay it's 6 steps and if you do it enough times it will 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.