Properties of Conjugate:Complex Number

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Polar form or Trigonometric form of Complex Number:
OA = |z|\cos \theta = x

 OP = |z|\sin \theta = y

 Since, z = x + iy

 \Rightarrow z = |z|(\cos \theta + i\sin \theta )

 \cos \theta = 1 - \frac{{{\theta ^2}}}{{2!}} + \frac{{{\theta ^4}}}{{4!}} - ....................

 or, \cos \theta = 1 + \frac{{{{(i\theta )}^2}}}{{2!}} + \frac{{{{(i\theta )}^4}}}{{4!}} + ..............

 \sin \theta = \theta - \frac{{{\theta ^3}}}{{3!}} + \frac{{{\theta ^5}}}{{5!}} - ....................

 or, i\sin \theta = i\theta + \frac{{{{(i\theta )}^3}}}{{3!}} + \frac{{{{(i\theta )}^5}}}{{5!}} + .............

 Therefore,

 z = |z|\left( {1 + i\theta + \frac{{{{(i\theta )}^2}}}{{2!}} + \frac{{{{(i\theta )}^3}}}{{3!}} + .......} \right)

 z = |z|{e^{i\theta }}          (Euler form)

 z = |z|cis\theta   ; cis\theta = \cos \theta + \sin \theta



Some Points about 'i':
1. If we square any number we will get a positive number.

 But, imagine there is an imaginary number 'i' which when square gives a negative number -1.

 {i^2} = - 1

 then, i = \sqrt { - 1}

Remember,

 \sqrt { - 3} \sqrt { - 4} \ne \sqrt {( - 3)( - 4)}

but, \sqrt { - 3} \times \sqrt { - 4} = \sqrt { - 1} \sqrt 3 \times \sqrt { - 1} \sqrt 4

or,\sqrt { - 3} \sqrt { - 4} = i\sqrt 3 \times i\sqrt 4 = {i^2}\sqrt 3 \times \sqrt 4

or, \sqrt { - 3} \times \sqrt { - 4} = - \sqrt 3 \times \sqrt 4   ( since {i^2} = - 1)


2. i = i
{i^2} = - 1
{i^3} = {i^2}i = - i
{i^4} = {i^2}{i^2} = ( - 1) \times ( - 1) = 1
{i^5} = {i^4}{i^{}} = i

 {i^6} = {i^4}{i^2} = 1 \times ( - 1) = - 1

 {i^7} = {i^4}{i^3} = 1 \times ( - i) = - i

 {i^8} = {i^4}{i^4} = 1

 ....................................................
...............................................

 Therefore, power of 'i' repeats its values and can attain only four values i.e. i, -1, -i and 1

 In general,
{i^{4n}} = 1 , {i^{4n + 1}} = i , {i^{4n + 2}} = - 1 , {i^{4n + 3}} = - i  , where n \in I
Reciprocal of i :

 \frac{1}{i} = \frac{1}{i} \times \frac{i}{i} = \frac{i}{{{i^2}}} = - i



Properties of Conjugate:

We know,

If z = x + iy

Conjugate of 'z',

\overline z = x - iy



1. \overline {\overline z } = z


2. z + \overline z = 2{{\rm Re}\nolimits} (z)

Here Re(z) represents the Real Part of complex number z.

Any complex number is purely imaginary if and only if z + \overline z = 0

Purely imaginary means Real part of complex number is zero i.e. Re(z) = 0 ( e.g. 2, 3, -4, -5/19 etc)

i.e. if z + \overline z = 0 then number is purely imaginary 
and if number is purely imaginary then z + \overline z = 0

Pure{{\rm Im}\nolimits} \Leftrightarrow z + \overline z = 0



3. z - \overline z = 2{{\rm Im}\nolimits} (z)

 Here Im(z) represents the Imaginary Part of complex number z.

 Pure{{\rm Re}\nolimits} \Leftrightarrow z - \overline z = 0 

 Purely Real means Imaginary part of complex number is zero i.e. Im(z) = 0 ( e.g. 2i, 8i, -90i etc )

 Any complex number is purely imaginary if and only if z - \overline z = 0

e.g. Show that z = {\left( {\frac{{2 - 3i}}{{4 + 5i}}} \right)^{2000}} + {\left( {\frac{{2 + 3i}}{{4 - 5i}}} \right)^{2000}} is purely Real ?
Since, z = {\left( {\frac{{2 - 3i}}{{4 + 5i}}} \right)^{2000}} + {\left( {\frac{{2 + 3i}}{{4 - 5i}}} \right)^{2000}}

 To find the conjugate of z we replace i from -i,

 Therefore,

 \overline z = {\left( {\frac{{2 + 3i}}{{4 - 5i}}} \right)^{2000}} + {\left( {\frac{{2 - 3i}}{{4 + 5i}}} \right)^{2000}}

 so,z - \overline z = 0

 Therefore z is purely Real.

4. \overline {{z_1} \pm {z_2}} = \overline {{z_1}} \pm \overline {{z_2}}


5. \overline {\left( {\frac{{{z_1}}}{{{z_2}}}} \right)} = \frac{{\overline {{z_1}} }}{{\overline {{z_2}} }}
6. \overline {{z_1}{z_2}} = \overline {{z_1}} \times \overline {{z_2}}


IMP
7. {z_1}\overline {{z_2}} + {z_2}\overline {{z_1}} = 2{{\rm Re}\nolimits} ({z_1}\overline {{z_2}} ) = 2{{\rm Re}\nolimits} ({z_2}\overline {{z_1}} )

 Explanation, 
Since \overline {\overline z } = z

 Therefore,  {z_2}\overline {{z_1}} = \overline {{z_1}\overline {{z_2}} }

And, z + \overline z = 2{{\rm Re}\nolimits} (z)

 Therefore,{z_1}\overline {{z_2}} + {z_2}\overline {{z_1}} = {z_1}\overline {{z_2}} + \overline {{z_1}\overline {{z_2}} } = 2{{\rm Re}\nolimits} ({z_1}\overline {{z_2}} )


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