Let z be the complex number defined as
$z=x+iy$
The conjugate of z is defined as
$\bar{z}=x-iy$
So by defination of Conjugate of any complex number is obtained by replacing i with -i
Proof
Let z_{1} = a + ib and z_{2} = c + id
then $\bar{z_1} = a - ib$ and $\bar{z_2} = c - id$
Now, $\bar{z_1 - z_2} = \bar{a + ib - c -id} = \bar{a - c +i(b - d)}=a + c - i(b - d)$
Also $\bar{z_1} - \bar{z_2} =a - ib - c + id=a - c - i(b - d)$
Hence Proved
Proof
Let z_{1} = a + ib and z_{2} = c + id
then $\bar{z_1} = a - ib$ and $\bar{z_2} = c - id$
Now, $\bar{z_1 + z_2} = \bar{a + ib + c + id} = \bar{a + c + i(b + d)}=a + c - i(b + d)$
Also $\bar{z_1} + \bar{z_2} =a - ib +c - id=a + c - i(b + d)$
Hence Proved
Proof
Let z_{1} = a + ib and z_{2} = c + id
then $\bar{z_1} = a - ib$ and $\bar{z_2} = c - id$
Now, $\bar{z_1 z_2} = \bar{ac - bd + i(ad + bc)} = ac - bd + i(ad + bc)$
when $z_{2}$ is not zero Proof
Let z_{1}/z_{2} = z_{3}
then z_{1}=z_{2}z_{3}
$\bar{z_1}=\bar{z_2 z_3}= \bar{z_2}\bar{z_3}$
or $\bar{z_3}= \frac {\bar{z_1}} {\bar{z_2}}$ or $\bar {\frac {z_1}{z_2}} =\frac {\bar{z_1}} {\bar{z_2}}$
Hence Proved
Example:
z=2-3i
Then
$\bar{z}=2+3i$
Modulus Of complex Number
The module of a complex number
$z=x+iy$
is defined as |z|
|z|=$\sqrt{(x^2+y^2)}$
Clearly |z| >= 0
Properties of Modulus of Complex Number
For $z=x+iy$
|z| =0 then it mean x=y=0
$|z|=|\bar{z}|=|-z|$
$z\bar{z}=|z|^{2}$
Proof
Let z = a + ib, then $\bar{z} = a - ib$
Now LHS
$z\bar{z} = (a + ib)(a - ib)=a^2 -i^2b^2 =a^2+b^2$
RHS
$|z|^{2}= a^2+b^2$
Hence Proved
$|z_{1}z_{2}|=|z_{1}||z_{2}|$
$|\frac{z_{1}}{z_{2}}|=\frac{|z_{1}|}{|z_{2}|}$ when $z_{2}$ is not zero
The proof of the above properties are quite self explanatory
Example
z=3-4i
Then
|z|=5
Reciprocal or Multiplicative Inverse of Complex Number using Complex Conjugate
z= x+iy
Then
$\frac{1}{z}=\frac{\bar{z}}{|z|^{2}}$
2) Solving the equation
(1+2i)z=(1-i) Solution:
$z=\frac{1-i}{1+2i}$
Dividing the conjugate for denominator
$z=[\frac{1-i}{1+2i}][\frac{1-2i}{|1-2i}]$
$z=[\frac{1-2i-i+2}{5}]$
$z=[\frac{3-i}{5}]$
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