Question

The complex number i: If we define \(i=\sqrt{-1}\), show that \(i^{2}=-1,i^{3}=-i, and i^{4}=1\).

Short answer

It is shown that \(i^{2}=-1,i^{3}=-i, and i^{4}=1\).

Step-by-step solution

Step 1. Given Information

It is given that \(i=\sqrt{-1}\). The 'i' is known as an imaginary number and it is used to represent a complex number that is non-real.

Step 2. Show

As it is given that \(i=\sqrt{-1}\) so if we square on both sides we get,

\(i=\sqrt{-1}\)

\(i^{2}=\left ( \sqrt{-1} \right )^{2}\)

\(i^{2}=-1\) ......(i)

The imaginary number, when multiplied by itself, gives a negative value.

If we multiplied \(i\) on both sides of equation (i) we get,

\(i^{2}=-1\)

\(i\times i^{2}=-1\times i\)

\(i^{3}=-i\).......(ii)

Now, if we multiplied \(i\) on both sides of equation (ii) we get,

\(i^{3}=-i\)

\(i\times i^{3}=-i\times i\)

\(i^{4}=-i^{2}\)

\(i^{4}=1\)