# De-Moivre's Theorem

**DE - MOIVRE’S THEOREM **

Let \(z = r{e^{i\theta }}\)

\(\begin{align}{} {\bf{(a)}}\quad\text{ If } n\text{ is an integer},\; {z^n} &= {(r{e^{i\theta }})^n} = {r^n}{e^{in\theta }}\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= {r^n}(\cos n\theta + i\sin n\theta ) \end{align}\)

\(\begin{align}{\bf{(b)}}\quad\text{If } n \text{ is a non-integer rational number, say of the form }\frac{p}{q},\end{align}\)

\[\begin{align}&{z^n} = {(r{e^{i\theta }})^{\frac{p}{q}}}\\\,\,\,\,\, &\quad= {r^{p/q}}{e^{\frac{{ip\theta }}{q}}} = {r^{p/q}}(\cos \frac{{p\theta }}{q} + i\sin \frac{{p\theta }}{q})\end{align}\]

is one of the values of \({z^n}\) . There will be actually multiple values of \({z^n}\) . (How to obtain those multiple values will be discussed subsequently).

We will now discus show to evaluate the \({n^{th}}\) roots of an arbitrary complex number in the manner described above.

\[\begin{align}&\;\;\;Let\,\,\,z = r{e^{i(2p\pi + \theta )}}\\ &\Rightarrow \,\,\,{z^{\frac{1}{n}}} = {\{ r{e^{i(2p\pi + \theta )}}\} ^{\frac{1}{n}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\quad\;= {r^{1/n}}{e^{i\left( {\frac{{2p\pi + \theta }}{n}} \right)}}\end{align}\]

This will have *\(n\)* unique values given by *\(n\)* successive integral values of \(p.\) We take *\(p\)* from the set \({0, 1, 2...(n–1)}.\)

The *\(n\)* values that we obtain are listed out below:

\[\begin{align}&{r^{1/n}}{e^{\frac{{i\theta }}{n}}},\,\,{r^{1/n}}{e^{i\left( {\frac{{2\pi }}{n} + \frac{\theta }{n}} \right)}},\,\,{r^{1/n}}{e^{i\left( {\frac{{4\pi }}{n} + \frac{\theta }{n}} \right)}}...,\,{r^{1/n}}{e^{i\left( {\frac{{2p\pi }}{n} + \frac{\theta }{n}} \right)}}...{r^{1/n}}{e^{i\left( {\frac{{2(n - 1)\pi }}{n} + \frac{\theta }{n}} \right)}}\end{align}\]

These *\(n\)* values of \({z^{1/n}}\) are termed the \({n^{th}}\) roots of \(z.\) How will they lie on a plane? Notice that the angle between any two successive roots \(\begin{align}{z_i}\,{\rm{ and }}\,{z_{i + 1}}{\rm{ is }}\frac{{2\pi }}{n}.\end{align}\) Thus, the \({n^{th}}\) roots will lie on a circle of radius \({r^{1/n}}\) and will be “evenly spaced out”; the angle between any two successive roots being \(\begin{align}\frac{{2\pi }}{n}\end{align}\).

Now we consider a special case, the \({n^{th}}\) roots of unity (1). In other words, we want the solutions to the equation:

\[{z^n} = 1\]

The Euler’s form of 1 is \({e^{i0}}.\) Therefore,

\[\begin{align}&z = {e^{i\frac{{(2p\pi + 0)}}{n}}}\\\,\,\, &\;= {e^{\frac{{i2p\pi }}{n}}}\end{align}\]

The *\(n\)* different values are

\[\begin{align}&1,\,\,{e^{\frac{{i2\pi }}{n}}},\,\,{e^{\frac{{i4\pi }}{n}}}...,\,\,{e^{\frac{{i2(n - 1)\pi }}{n}}}\end{align}\]

If we let \({e^{i2\pi /n}} = \alpha \) , the *n* root are \(1,\,\,\alpha ,\,\,{\alpha ^2}...\,\,{\alpha ^{n - 1}}\)

These *\(n\)* roots will be evenly spaced out, lying on a circle of radius 1 centred at the origin..

From symmetry, notice that the sum of these *\(n\)* roots (vectors) will be 0.

\[1 + \alpha + {\alpha ^2} + ... + {\alpha ^{n - 1}} = 0\]

This can be proved analytically too:

Since \(\alpha \)* *is one of the \({n^{th}}\) roots of 1, we have

\[\begin{align}&\quad\quad\;{\alpha ^n} = 1\\ &\Rightarrow \,\,\,{\alpha ^n} - 1 = 0\\ &\Rightarrow \,\,\,(\alpha - 1)({\alpha ^{n - 1}} + {\alpha ^{n - 2}} + ... + \alpha + 1) = 0\end{align}\]

Since \(\alpha \ne 1\) , we get:

\[{\alpha ^{n - 1}} + {\alpha ^{n - 2}} + ... + \alpha + 1 = 0\]

Now we evaluate the product of these *n* roots:

\[\begin{align}&P = 1 \cdot \alpha \cdot {\alpha ^2}...{\alpha ^{n - 1}}\\ &\;\;= {\alpha ^{1 + 2 + 3 + ... + n - 1}}\\ &\;\;= {\alpha ^{\frac{{n(n - 1)}}{2}}}\end{align}\]

Since \(\begin{align}&\alpha = {e^{i2\pi /n}},P = {e^{\frac{{i2\pi }}{n} \cdot \left( {\frac{{n(n - 1)}}{2}} \right)}} = {e^{i\pi (n - 1)}}\end{align}\) .

Now observe that if *\(n\)* is odd, *\(P\)* is 1 while if *\(n\)* is even, *\(P\)* is –1:

\[P = \,\,\left\{ {\begin{array}{*{20}{c}}1&{{\rm{if \,\,}}n{\rm{ \,\,is \,\,odd}}}\\{ - 1}&{{\rm{if \,\,}}n{\rm{ \,\,is \,\,even}}}\end{array}} \right\}\]

Lets apply this discussion to some particular values of \(n:\)

**(i) ** \(\fbox{n = 2}\)

We want the square roots of unity, or, the solutions to

\[{z^2} = 1\]

One of the roots is 1. Where can we symmetrically place the other root? Obviously, at –1.

**(ii) ** \(\fbox{n = 3}\) ** *** *

We want to solve

\[{z^3} = 1\]

One of the roots is 1. The other roots can be placed symmetrically if they are at angles, \(\begin{align}&\frac{{2\pi }}{3}{\rm{ and }}\frac{{4\pi }}{3}\end{align}\) as shown below:

The two other roots are:

\[\omega = {e^{i\,2\pi /3}} = - \frac{1}{2} + \frac{{i\sqrt 3 }}{2},\,\,{\omega ^2} = {e^{i\,4\pi /3}} = - \frac{1}{2} - \frac{{i\sqrt 3 }}{2}\]

Notice that \(w\,\,and\,\,w{^2}\) are conjugates of each other. Also, \(1 + \omega + {\omega ^2} = 0\) and \(1 \cdot \omega \cdot {\omega ^2} = {\omega ^3} = 1\)

**(iii) \(\fbox{n = 4}\)**

We want to solve

\[{z^4} = 1\]

The four roots can be symmetrically placed as shown below:

Note that \(1 + i + ( - 1) + ( - i) = 0\,\,{\rm{and }}1 \cdot (i) \cdot ( - 1) \cdot ( - i) = - 1\) .

Higher order roots can be similarly evaluated. You must understand carefully the geometrically significance of the \({n^{th}}\) roots. Let us take one of the cube roots of unity for this purpose, say \({\omega ^2}\) .

If you cube \({\omega ^2}\) , you are essentially rotating this vector. The argument of \({\omega ^2}\) is 240° ; when you cube \({\omega ^2}\) , you will rotate by 240° \(({\rm{for\,\, }}{\omega ^2} \times {\omega ^2}) + {\rm{another\,\,}} 240^\circ ({\rm{for\,\, }}{\omega ^2} \times {\omega ^2} \times {\omega ^2})\) . Hence, \({\omega ^2}\) , when cubed, will become the vector 1.

Thus,

\[{({\omega ^2})^3} = 1\]

Similarly, observe one of the fourth roots of unity, say \(i.\) The argument of *\(i\)* is 90°. When you raise *\(i\)* to power 4, you are essentially rotating the vector *\(i\)* by 90°(for\( i × i\)) + 90°(for \(i ×i ×i\)) + 90°(for \(i×i×i×i\)). Therefore, *\(i\)* when raised to power four will become the vector 1.

Thus,

\[{i^4} = 1\]

This discussion should make you realise that when looking for the \({n^{th}}\) roots of unity, you are looking for vectors which when rotated by a certain fixed angle \((2\pi /n)\) a particular number \((n – 1)\) of times, give the vector 1.