# Why we don't like i in the denominator?

#### babaliaris

I forgot why we learned at school to bring i in the top of a fraction (in the numerator).

Why do we do that? Is it because we can clearly see the complex number in this way?

#### idontknow

The general method works as $$\displaystyle z=a+bi$$.
For example $$\displaystyle z=\frac{1}{i}=\frac{i}{i^2 } = \frac{i}{-1}=-i$$.
Now $$\displaystyle a=0$$ and $$\displaystyle b=-1$$.

#### DarnItJimImAnEngineer

Yeah, that's pretty much it; makes it easy to separate the real and imaginary parts. Another benefit is it puts complex numbers in a sort of standardised form that makes it easier to compare them -- the same way we do with other radicals.

2 people

#### SDK

There is no mathematical reason whatsoever to prefer $i$ in the numerator of a fraction vs denominator. Sounds like you had a sadistic professor.

#### Maschke

There is no mathematical reason whatsoever to prefer $i$ in the numerator of a fraction vs denominator. Sounds like you had a sadistic professor.
I agree with half of this. There's no mathematical reason whatsoever. But it's the standard convention, just like putting radicals in the numerator. Conventions do have meaning, such as for example driving on the right or left according to your country's laws. There's no reason for it, but you get in big trouble if you don't do what everyone else does.

#### romsek

Math Team
but you get in big trouble if you don't do what everyone else does.
they'll look at the denominator and poke your i out!

Maschke

#### DarnItJimImAnEngineer

There's no mathematical reason whatsoever.
I don't know about mathematical reasons, but there are definitely practical reasons. We use the poles of transfer functions to determine stability of systems. If the real part of a pole is negative, it is stable in that mode. The more negative it is, the more stable it is.
If you find a pole at $\displaystyle \frac{3\pm 5i}{-1\pm i}$, it is far from obvious whether the real part is positive or negative. If you write it as $1\pm 4i$, it is much more evident that you have an instability.

Maschke