Euler's Formula And De Moivre's Theorem

Euler’s formula

Euler’s formula states that ‘For any real number

Let z be a non zero complex number; we can write

Multiplying a complex number

De Moivre’s theorem

It states that for any integer

This can be easily proved using Euler’s formula as shown below.

We know that,

Therefore,

If any complex number satisfies the equation

Fundamental theorem of algebra says that, an equation of degree

To find the values of

We have,

From (1),

By De Moivre’s theorem,

For example; if

We know that,

Let

When

Therefore,

• Sum of
  \(\begin{array}{l}n^{th}\end{array} \)
  roots of unity is,
  \(\begin{array}{l}1~+~ω~+~ω^2~+~ω^3~+~⋯~+~ω^{n~-~1}\end{array} \)
  It is geometric series having first term 1 and common ratio
  \(\begin{array}{l}ω\end{array} \)
  .By using sum of
  \(\begin{array}{l}n\end{array} \)
  terms of a G.P,
  \(\begin{array}{l}1~+~ω~+~ω^2~+~ω^3~+~⋯~+~ω^{n~-~1}\end{array} \)
  =
  \(\begin{array}{l}\frac{1~-~ω^n}{1~-~ω}\end{array} \)
  Since
  \(\begin{array}{l}ω\end{array} \)
  is
  \(\begin{array}{l}n^{th}\end{array} \)
  root of unity,
  \(\begin{array}{l}ω^n\end{array} \)
  =
  \(\begin{array}{l}1\end{array} \)
  Therefore,
  \(\begin{array}{l}1~+~ω~+~ω^2~+~ω^3~+~⋯~+~ω^{n~-~1}\end{array} \)
  =
  \(\begin{array}{l}0\end{array} \)

Cube roots of unity:

We know that

Therefore, cube roots of unity are

Sum of the cube roots of the unity,

Product of cube roots of the unity,

Example:

Roots of the equation are

Values of

Therefore,

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