Conjugates Of Complex Numbers

CONJUGATION:

Consider a complex number \(z = x + iy .\) Where do you think will the number \(x - iy\) lie? A little thinking will show that it will be the exact mirror image of the point \(z\), in the x-axis mirror.

Such a number is given a special name. It is called the conjugate of \(z\) and represented as \(\bar z\). Therefore,

if \(z = x + iy,\) then  \(\bar z = x - iy\) .

It turns out the notion of a conjugate is, though simple, very useful. We will put the conjugate to a lot of use in this chapter.

Let us now see some of the properties of the conjugate.

(1)  \(\left| z \right| = \left| {\bar z} \right| \text{ and } \arg (\bar z) =  - \arg (z)\) .

The validity of these two relations should be obvious from the figure below:

(2) \(\overline {{z_1} \pm {z_2}}  = {\bar z_1} \pm {\bar z_2}\)

These relations mean that the operation of conjugation is distributive over addition (and subtraction). You are urged to verify this by considering two arbitrary complex numbers \({z_1} = {x_1} + i{y_1}\)  and  \({z_2} = {x_2} + i{y_2}\) .

 

(3)  \(\begin{align}\overline {{z_1}{z_2}}  = {\bar z_1}{\bar z_2};\,\,\,\,\overline {\left( {\frac{{{z_1}}}{{{z_2}}}} \right)}  = \frac{{{{\bar z}_1}}}{{{{\bar z}_2}}}\end{align}\)  

These relations imply that conjugation is distributive over multiplication and division too. Lets prove this for multiplication.

Let \({z_1} = {x_1} + i{y_1}\)  and  \({z_2} = {x_2} + i{y_2}\)  .

Then,   \(\overline {{z_1}{z_2}}  = \overline {({x_1} + i{y_1})({x_2} + i{y_2})} \) .

\( = \overline {({x_1}{x_2} - {y_1}{y_2}) + i({x_1}{y_2} + {x_2}{y_1})} \) .

\( = ({x_1}{x_2} - {y_1}{y_2}) - i({x_1}{y_2} + {x_2}{y_1})\)

Now,   \({\bar z_1}{\bar z_2} = \overline {({x_1} + i{y_1})} \,\,\,\overline {({x_2} + i{y_2})} \) .

\( = ({x_1} - i{y_1})({x_2} - i{y_2})\) .

\( = ({x_1}{x_2} - {y_1}{y_2}) - i({x_1}{y_2} + {x_2}{y_1})\)

We see that \(\overline {{z_1}{z_2}}  = {\bar z_1}{\bar z_2}\) . The proof for division is similar.

(4)  \({\left| z \right|^2} = z\bar z\)

This is one of the most important relations to be used in this chapter. The proof is very straightforward.

Let \(z = x + iy.\)

\[\begin{align}z\bar z &= (x + iy)(x - iy)\\&= {x^2} - {i^2}{y^2}\\&= {x^2} + {y^2}\\&= {(\sqrt {{x^2} + {y^2}} )^2}= {\left| z \right|^2}\end{align}\]

(5)  \(z + \bar z = 2{\mathop{\rm Re}\nolimits} (z);\,\,z - \bar z = 2i{\mathop{\rm Im}\nolimits} (z)\) . \({\mathop{\rm Re}\nolimits} (z)\)  denotes the real part of z and Im(z) denotes the imaginary part of \(z.\)

This property is again obvious; just consider  \(z = x + iy\)  to deduce it.

Example- 6

Simply the following

(a)  \({\left| {{z_1} \pm {z_2}} \right|^2}\)                    (b)  \({\left| {{z_1} + {z_2}} \right|^2} + {\left| {z{{\kern 1pt} _1} - {z_2}} \right|^2}\)

 Solution: (a) Let us simplify  \({\left| {{z_1} + {z_2}} \right|^2}\)

\[\begin{align}{\left| {{z_1} + {z_2}} \right|^2} &= ({z_1} + {z_2})(\overline {{z_1} + {z_2}} ) \qquad \qquad \text{(property 4)}\\&= ({z_1} + {z_2})({\bar z_1} + {\bar z_2}) \qquad \qquad \text{(property 2)}\\& = {z_1}{\bar z_1} + {z_2}{\bar z_2} + {z_1}{\bar z_2} + {\bar z_1}{z_2}\\&= {\left| {{z_1}} \right|^2} + {\left| {{z_2}} \right|^2} + {z_1}{\bar z_2} + {\bar z_1}{z_2}\end{align}\]

Now notice that \({\bar z_1}{z_2}\)  can be written as  \(\overline {{z_1}{{\bar z}_2}} \) by virtue of property (3) \(\{ \overline {{z_1}{{\bar z}_2}}  = {\bar z_1}{\overline{\overline {\,z\,}} _2} = {\bar z_1}{z_2}\} \). Therefore,  \({z_1}{\bar z_2} + {\bar z_1}{z_2}\)  becomes  \({z_1}{\bar z_2} + \overline {{z_1}{{\bar z}_2}} \) which is equal to \(2{\mathop{\rm Re}\nolimits} ({z_1}{\bar z_2})\) by property (5).

Hence,

\[{\left| {{z_1} + {z_2}} \right|^2} = {\left| {{z_1}} \right|^2} + {\left| {{z_2}} \right|^2} + 2{\mathop{\rm Re}\nolimits} ({z_1}{\bar z_2})\]

Similarly,

\[{\left| {{z_1} - {z_2}} \right|^2} = {\left| {{z_1}} \right|^2} + {\left| {{z_2}} \right|^2} - 2{\mathop{\rm Re}\nolimits} ({z_1}{\bar z_2})\]

(b) The solution to this part is obtained by simply adding the two expressions obtained in part (a):

 \[{\left| {{z_1} + {z_2}} \right|^2} + {\left| {{z_1} - {z_2}} \right|^2} = 2({\left| {{z_1}} \right|^2} + {\left| {{z_2}} \right|^2})\]

Example- 7

If  \({z_1},\,\,{z_2},\,\,{z_3}\,\) represent the vertices of an equilateral triangle, show that \(\begin{align}\frac{1}{{{z_1} - {z_2}}} + \frac{1}{{{z_2} - {z_3}}} + \frac{1}{{{z_3} - {z_1}}} = 0\end{align}\) .

Solution:

We know that the sides of an equilateral triangle are equal. Using this for the triangle above, we get:

\[\left| {{z_1} - {z_2}} \right| = \left| {{z_2} - {z_3}} \right| = \left| {{z_3} - {z_1}} \right| = l\]

Now, we square this to get

\[{\left| {{z_1} - {z_2}} \right|^2} = {\left| {{z_2} - {z_3}} \right|^2} = {\left| {{z_3} - {z_1}} \right|^2} = {l^2}\]

or \(({z_1} - {z_2})({\bar z_1} - {\bar z_2}) = ({z_2} - {z_3})({\bar z_2} - {\bar z_3}) = ({z_3} - {z_1})({\bar z_3} - {\bar z_1}) = {l^2}\) (properties 5 and 2)

\[\begin{align}\Rightarrow \qquad \frac{1}{{{z_1} - {z_2}}} = \frac{{{{\bar z}_1} - {{\bar z}_2}}}{{{l^2}}};\,\,\,\,\,\frac{1}{{{z_2} - {z_3}}} = \frac{{{{\bar z}_2} - {{\bar z}_3}}}{{{l^2}}};\,\,\,\,\,\frac{1}{{{z_3} - {z_1}}} = \frac{{{{\bar z}_3} - {{\bar z}_1}}}{{{l^2}}}\end{align}\]

Adding these three, we get the desired result.

Example- 8

Prove that if the sum and product of two non-real complex numbers are real, they must be the conjugates of each other.

Solution: Let \({z_1} = {x_1} + i{y_1} \,\,\,\,\text{and} \,\,\,\,{z_2} = {x_2} + i{y_2} \,\,\,\,\,\,\,\,\text{where }{y_1},{y_2} \ne 0\) .

It is given that   \({z_1} + {z_2}\)  and  \({z_1}{z_2}\) are purely real, i.e.,

\[{\mathop{\rm Im}\nolimits} ({z_1} + {z_2}) = {\mathop{\rm Im}\nolimits} ({z_1}{z_2}) = 0`\]

\[\begin{align}&{\mathop{\rm Im}\nolimits} ({z_1} + {z_2}) = {\mathop{\rm Im}\nolimits} ({z_1}{z_2}) = 0`\\&\Rightarrow {y_1} + {y_2} = 0\;\,\,\text{and} \;\;{x_1}{y_2} + {x_2}{y_1} = 0\\&\Rightarrow {y_1} = - {y_2}\,\,\;\text{and}\;\; {x_1}{y_2} - {x_2}{y_2} = 0\\& \Rightarrow {x_1} = {x_2}\end{align}\]

Thus, \({z_1} = {x_1} + i{y_1} \;  and \;  {z_2} = {x_2} + i{y_2} = {x_1} - i{y_1}\)  or in other words,

 \({z_1} \;  \text{and} \;  {z_2}\) are conjugates of each other.

Example- 9

If  \({\left| {{z_1} + {z_2}} \right|^2} = {\left| {{z_1}} \right|^2} + {\left| {{z_2}} \right|^2}\)  for two arbitrary non-zero complex numbers, show that \(\begin{align}\frac{{{z_1}}}{{{z_2}}}\end{align}\) is purely imaginary.

Solution:  From example-6,\({\left| {{z_1} + {z_2}} \right|^2}\)  can be written as  \({\left| {{z_1}} \right|^2} + {\left| {{z_2}} \right|^2} + 2{\mathop{\rm Re}\nolimits} ({z_1}{\bar z_2})\)

\( \Rightarrow \)  \({\left| {{z_1}} \right|^2} + {\left| {{z_2}} \right|^2} + 2{\mathop{\rm Re}\nolimits} \,({z_1}{\bar z_2}) = {\left| {{z_1}} \right|^2} + {\left| {{z_2}} \right|^2}\)

\( \Rightarrow \) \({\mathop{\rm Re}\nolimits} ({z_1}{\bar z_2}) = 0\)

Since the real part of \({z_1}{\bar z_2}\) is 0 , it must be purely imaginary. This means that it lies on the imaginary axis or its argument must be \(\begin{align}\frac{\pi }{2}{\rm{or}}\frac{{ - \pi }}{2}\end{align}\) .

\[ \begin{align}&\Rightarrow \qquad Arg ({z_1}{\bar z_2}) =  \pm \frac{\pi }{2}\\&\Rightarrow \qquad Arg ({z_1}) + Arg({\bar z_2}) =  \pm \frac{\pi }{2} \qquad
 \text{[Arguments of complex number when they are multiplied]}\\
&\Rightarrow \qquad {\rm{Arg}}({z_1}) - {\rm{Arg}}({z_2}) =  \pm \frac{\pi }{2} \qquad [Arg (\bar z) = -Arg(z)]\\&\Rightarrow \qquad Arg \left( {\frac{{{z_1}}}{{{z_2}}}} \right) =  \pm \frac{\pi }{2} \qquad  \qquad \;\;\; \text{[Argument of complex numbers subtract when they are divided]}\end{align}\] 

This shows that  \(\begin{align}\frac{{{z_1}}}{{{z_2}}}\end{align}\)  lies on  the imaginary axis; therefore, it is purely imaginary.

Example- 10

If  \({z_1},{z_2}....{z_n}\) are complex numbers and \(\left| {{z_1}} \right| = \left| {{z_2}} \right| = ... = \left| {{z_n}} \right| = 1,\) show that

\[\left| {{z_1} + {z_2} + ... + {z_n}} \right| = \left| {\frac{1}{{{z_1}}} + \frac{1}{{{z_2}}} + ... + \frac{1}{{{z_n}}}} \right|\]

Solution: The information that all the moduli are 1 and the form of the expression we need to obtain hint that we should consider the term  \({\left| {{z_i}} \right|^2}\)  :

\[\begin{align}&{\left| {{z_i}} \right|^2} = {z_i}{\bar z_i} = 1\\&\Rightarrow {z_i} = \frac{1}{{{{\bar z}_i}}} = \overline {\left( {\frac{1}{{{z_i}}}} \right)}\\&\Rightarrow \sum\limits_{i = 1}^n \left[ {\left| z \right|} = {\left| {\bar z} \right|} \right]{{z_i}}  = \sum\limits_{i = 1}^n {\overline {\left( {\frac{1}{{{z_i}}}} \right)} } \\&= \overline {\left( {\sum\limits_{i = 1}^n {\frac{1}{{{z_i}}}} } \right)} \qquad \quad \text{[Conjugation distributes over addition]}\\&\Rightarrow \left| {\sum\limits_{i = 1}^n {{z_i}} } \right| = \left| {\overline {\left( {\sum\limits_{i = 1}^n {\frac{1}{{{z_i}}}} } \right)} } \right|\\&= \left| {\left( {\sum\limits_{i = 1}^n {\frac{1}{{{z_i}}}} } \right)} \right|\end{align}\] 

This is the result we needed to obtain.