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.
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