De-Moiver's Theorem: Complex Number

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De-Moiver's Theorem:-

if n \in Q then,

{(\cos \theta + i\sin \theta )^n} =

CASE 1st: If n is an integer then,

{(\cos \theta + i\sin \theta )^n} = \cos n\theta + i\sin n\theta


CASE 2nd: If n is a rational number ( different from integers)

i.e. n = p/q where p,q belong to integers, q \ne 0 and p,q have no common divisor then

{(\cos \theta + i\sin \theta )^n} has q distinct values and one of these is equal to \cos n\theta + i\sin n\theta .



Steps of root finding of a complex number 'z' :

Step 1: Convert 'z' into the 'Polar form' where \theta  is the Principal value of the arg.z.

i.e. z = |z|(\cos \theta + i\sin \theta )

Step 2: Generalise \theta ,

i.e. z = |z|\left\{ {\cos (2k\pi + \theta ) + i\sin (2k\pi + \theta )} \right\} , k belongs to integers 

Step 3: The nth roots are given by,

{z^{1/n}} = |z{|^{1/n}}\left\{ {\cos (\frac{{2k\pi + \theta }}{n}) + i\sin (\frac{{2k\pi + \theta }}{n})} \right\}, where k = 0,1,2,3..........,n - 1

or, {z^{1/n}} = |z{|^{1/n}}{e^{i\left( {\frac{{2k\pi + \theta }}{n}} \right)}}



Ques: Find the 4th roos of z = - 1 + i\sqrt 3  ?



Sol: 

Step 1: Convert z = - 1 + i\sqrt 3  into polar form.

if z = x + iy and we want to convert it into polar form then we have to multiply and divide the sum of square of x and y.

therefore, z = 2\left( {\frac{{ - 1}}{2} + i\frac{{\sqrt 3 }}{2}} \right)

or, z = 2\left( {\cos \frac{{2\pi }}{3} + i\sin \frac{{2\pi }}{3}} \right)

Step 2: Generalize the \theta

therefore, z = 2\left\{ {\cos \left( {2k\pi + \frac{{2\pi }}{3}} \right) + i\sin \left( {2k\pi + \frac{{2\pi }}{3}} \right)} \right\}

Step 3: The 4th roots are given by,


{z^{1/4}} = {2^{1/4}}\left\{ {\cos \frac{{\left( {2k\pi + \frac{{2\pi }}{3}} \right)}}{4} + i\sin \frac{{\left( {2k\pi + \frac{{2\pi }}{3}} \right)}}{4}} \right\}, where k = 0,1,2,3

When , k = 0

{z_0} = {2^{1/4}}\left( {\cos \frac{{2\pi }}{{12}} + i\sin \frac{{2\pi }}{{12}}} \right) = {2^{1/4}}\left( {\cos \frac{\pi }{6} + i\sin \frac{\pi }{6}} \right)


similarly we can find the other roots i.e. {z_1},{z_2},{z_3}




NOTE: 
For nth roots:

{z^{1/n}} = |z{|^{1/n}}\left\{ {\cos (\frac{{2k\pi + \theta }}{n}) + i\sin (\frac{{2k\pi + \theta }}{n})} \right\}, where k = 0,1,2,3 ........ , n - 1

or, {z^{1/n}} = |z{|^{1/n}}{e^{i\left( {\frac{{2k\pi + \theta }}{n}} \right)}} , where k = 0,1,2,3 ........ , n - 1

for k = 0,  {z_0} = |z{|^{1/n}}{e^{i\frac{\theta }{n}}}

k = 1,   {z_1} = |z{|^{1/n}}{e^{i\frac{{2\pi + \theta }}{n}}}

or, {z_1} = |z{|^{1/n}}{e^{i\frac{\theta }{n}}}.{e^{i\frac{{2\pi }}{n}}}

 k = 2,    {z_2} = |z{|^{1/n}}{e^{i\frac{\theta }{n}}}.{e^{i\frac{{4\pi }}{n}}}

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k = n - 1,   {z_{n - 1}} = |z{|^{1/n}}{e^{i\frac{\theta }{n}}}.{e^{i\frac{{2(n - 1)\pi }}{n}}}

Therefore from above arguments, we can observe:
1. All these roots are in G.P. with common ratio of {e^{i\frac{{2\pi }}{n}}}

2. All these roots lie on a circle of radius |z{|^{1/n}} whose centre is at the Origin.

3. All these roots are the vertices of a Regular Polygon of n-sides



Cube roots of unity:
z = 1

or, {z^{}} = \cos {0^o} + i\sin {0^o}

or, z = \cos 2k\pi + i\sin 2k\pi

or, {z^{1/3}} = \cos \frac{{2k\pi }}{3} + i\sin \frac{{2k\pi }}{3},  where k = 0, 1, 2

or, {z^{1/3}} = {e^{i\frac{{2k\pi }}{3}}} , k = 0, 1 ,2

{z_0} = 1

{z_1} = {e^{i\frac{{2\pi }}{3}}} = {z_0}{e^{i\frac{{2\pi }}{3}}} =  - \frac{1}{2} + i\frac{{\sqrt 3 }}{2}

{z_2} = {e^{i\frac{{4\pi }}{3}}} = {z_1}{e^{i\frac{{2\pi }}{3}}} = - \frac{1}{2} - i\frac{{\sqrt 3 }}{2}

a) Sum of the roots are zero when the modulus of the roots is unity.

i.e.{z_0} + {z_1} + {z_2} = 0

or, 1 + {z_1} + {z_2} = 0

if {z_1} = w then {z_2} = {w^2}

therefore,1 + w + {w^2} = 0

b) Product of the roots of the unity = 1 when n is an odd number.

i.e. {w^3} = 1

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