Complex Numbers Set 4

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Example- 13

Prove these identities for an arbitrary complex number z:

(a)   \(\overline {({e^z})}  = {e^{\bar z}}\)   (b) \(\overline {(\ln z)}  = \ln \bar z\)   (c) \(\overline {(\cos z)}  = \cos \bar z\)       (d)\(\overline {(z_1^{{z_2}})}  = {\bar z_1}^{{{\bar z}_2}}\)   .

Solution:  Let   \(z=x+iy=r{{e}^{i\theta }}\)

\(\begin{align}(a)\;\overline{({{e}^{z}})}=\overline{({{e}^{x+iy}})}=\overline{{{e}^{x}}\cdot {{e}^{iy}}}&={{e}^{x}}\cdot {{e}^{-iy}}\\\qquad\qquad\qquad\qquad\quad&={{e}^{x-iy}}\\\qquad\qquad\qquad\quad\quad&={{e}^{{\bar{z}}}}\end{align}\)

You might find the operation of raising a real number to an arbitrary complex number sort of strange. But with time, you will realise that what we are doing is mathematically consistent and therefore makes sense.  For example, we can even determine sines and cosines of complex numbers! (as in part -c below)

 \(\begin{align}(b)\;&\overline {(\ln z)} = \overline {\ln (r{e^{i\theta }})} = \overline {(\ln r + \ln {e^{i\theta }})} \\   &\qquad\qquad\qquad\;\;\;= \overline {\ln r + i\theta } \\   &\qquad\qquad\qquad\;\;\;= \ln r - i\theta \\   &\qquad\qquad\qquad\;\;\;= \ln r + \ln {e^{ - i\theta }}\\  &\qquad\qquad\qquad\;\;\; = \ln r{e^{ - i\theta }}\\   &\qquad\qquad\qquad\;\;\;= \ln \bar z\end{align}\)

(c)  We can write cos z as \(\begin{align}\frac{{{e^{iz}} + {e^{ - iz}}}}{2}\end{align}\)

 \(\begin{align} \Rightarrow \,\,\,\overline {\cos z}  = \overline {\left( {\frac{{{e^{iz}} + {e^{ - iz}}}}{2}} \right)}  = \frac{{{e^{ - i\bar z}} + {e^{i\bar z}}}}{2}\end{align}\) (From part (a))

(d) \(z_1^{{z_2}}\) can be written as \({e^{{z_2}\ln {z_1}}}\)

\( \Rightarrow \,\,\,\overline {\left( {z_1^{{z_2}}} \right)}  = \overline {\left( {{e^{{z_2}\ln {z_1}}}} \right)}  = {e^{\overline {{z_2}\ln {z_1}} }} = {e^{{{\bar z}_2}\overline {(\ln {z_1})} }} = {e^{{{\bar z}_2}\ln {{\bar z}_1}}} = {\bar z_1}^{{{\bar z}_2}}\)

Example- 14

Find the largest and the smallest value of |z| if z satisfies

\[\begin{align}\left| {z + \frac{1}{z}} \right| = a\end{align}\]

Solution:   We apply the triangle inequality on \(\left| {z + \frac{1}{z}} \right|\):

\[a = \left| {z + \frac{1}{z}} \right| \ge \left| {\left| z \right| - \frac{1}{{\left| z \right|}}} \right|\]

\[\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\overbrace {\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\;\;\;\;\;\,\,\,}^{{\rm{Ineq A}}}\\ \Rightarrow \,\,\,\,\,\,\,\, - a \le \left| z \right| - \frac{1}{{\left| z \right|}} \le a\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\;\;\quad\underbrace {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\qquad}_{{\rm{Ineq B}}}\end{array}\]

Ineq A:\( \qquad\qquad\qquad\qquad {\left| z \right|^2} + a\left| z \right| - 1 \ge 0\)        

 \[ \Rightarrow \quad \quad \text{The roots of the quadratic expression are } \frac{{ - a \pm \sqrt {{a^2} + 1} }}{4}\\\Rightarrow \qquad    \text{ Since |z| > 0, we have the solution as}\left| z \right| \ge \frac{{ - a + \sqrt {{a^2} + 1} }}{4}\]

  Ineq B :    \(\qquad\qquad\qquad{\left| z \right|^2} - \left| z \right| - 1 \le 0\)

\[\begin{align}&\Rightarrow \qquad \text{  The roots of this quadratic expression are }\frac{{a \pm \sqrt {{a^2} + 1} }}{4}\\&\Rightarrow     \qquad   \left| z \right| \le \frac{{a + \sqrt {{a^2} + 1} }}{4}\end{align}\]

Thus,

   \[\frac{{ - a + \sqrt {{a^2} + 1} }}{4} \le \left| z \right| \le \frac{{a + \sqrt {{a^2} + 1} }}{4}\]

These are the required largest and smallest values, between which |z| can lie.

Example- 15

Assume that \({A_i}(i = 1,\,\,2...n)\) are the vertices of a regular polygon inscribed in a circle of radius unity. Find the value of \({\left| {{A_1}{A_2}} \right|^2} + {\left| {{A_1}{A_3}} \right|^2} + ... + {\left| {{A_1}{A_n}} \right|^2}\)

 Solution :  There is no loss of generality in assuming that one of the vertex, say Al, lies at the point 1.

Thus, \({A_1}{A_2} = \left| {1 - {e^{i2\pi /n}}} \right| = \left| {1 - \cos \frac{{2\pi }}{n} - i\sin \frac{{2\pi }}{n}} \right|\)

\[\begin{align}& =\left| {2{{\sin }^2}\frac{\pi }{n} - 2i\sin \frac{\pi }{n}\cos \frac{\pi }{n}} \right|\\\\ &= 2\sin \frac{\pi }{n}\left| {\sin \frac{\pi }{n} - i\cos \frac{\pi }{n}} \right|\\\\ &= 2\sin \,\,\pi /n\end{align}\]

Similarly,  \(\begin{align}{A_1}{A_r} = \left| {1 - {e^{i2(r - 1)\,\,\pi /n}}} \right| = 2\sin \frac{{(r - 1)\pi }}{n}\end{align}\)

Now,        \({\left| {{A_1}{A_2}} \right|^2} + {\left| {{A_1}{A_3}} \right|^2} + ... + {\left| {{A_1}{A_n}} \right|^2}\)

      \[\begin{align}& = 4{\sin ^2}\frac{\pi }{n} + 4{\sin ^2}\frac{{2\pi }}{n} + ... + 4{\sin ^2}\frac{{(n - 1)\pi }}{n}\\ &= 2\left\{ {\left( {1 - \cos \frac{{2\pi }}{n}} \right) + \left( {1 - \cos \frac{{4\pi }}{n}} \right) + ... + \left( {1 - \cos \frac{{2(n - 1)\pi }}{n}} \right)} \right\}\\ &= 2\left\{ {(n - 1) - \left( {\cos \frac{{2\pi }}{n} + \cos \frac{{4\pi }}{n} + ...\cos \frac{{2 \cdot (n - 1)\pi }}{n}} \right)} \right\}\end{align}\]

\[\begin{align}& = 2\left\{ {(n - 1) - ( - 1)} \right\} \left( \begin{array}{l}{\rm{Why? Because}}\sum\limits_{r = 0}^{n - 1} {{e^{i2\pi \,\,r/n}}}  = 0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\sum\limits_{r = 1}^{n - 1} {{e^{i2\pi \,\,r/n}}}  =  - 1\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,{\mathop{\rm Re}\nolimits} \left( {\sum\limits_{r = 1}^{n - 1} {{e^{i2\pi \,\,r/n}}} } \right) =  - 1\end{array} \right)\\\ &= {\rm{ }}2n\end{align}\]

Example- 16

If n is a positive integer, prove that \(\left| {{\mathop{\rm Im}\nolimits} ({z^n})} \right| \le n\left| {{\mathop{\rm Im}\nolimits} (z)} \right|{\left| z \right|^{n - 1}}\) .

Solution: \(\begin{align}{\mathop{\rm Im}\nolimits} ({z^n}) = \frac{{{z^n} - \overline {{z^n}} }}{{2i}} = \frac{{{z^n} - {{\bar z}^n}}}{{2i}}\end{align}\)

Similarly,     \(\begin{align}{\mathop{\rm Im}\nolimits} (z) = \frac{{z - \bar z}}{{2i}}\end{align}\)

\[\begin{align}& \Rightarrow \,\,\,\left| {\frac{{{\mathop{\rm Im}\nolimits} ({z^n})}}{{{\mathop{\rm Im}\nolimits} (z)}}} \right| = \left| {\frac{{{z^n} - {{\bar z}^n}}}{{z - \bar z}}} \right|\\\\ &\qquad\qquad\qquad= {\left| {\bar z} \right|^{n - 1}}\left| {\frac{{{{\left( {\frac{z}{{\bar z}}} \right)}^n} - 1}}{{\left( {\frac{z}{{\bar z}}} \right) - 1}}} \right|\\\\ &\qquad\qquad\qquad= {\left| {\bar z} \right|^{n - 1}}\left| {{\alpha ^{n - 1}} + {\alpha ^{n - 2}} + ...1} \right|  \left( {{\rm{\;\;\;\;\;we let }}\alpha  = \frac{z}{{\bar z}}} \right)\\\\ &\qquad\qquad\qquad\le {\left| {\bar z} \right|^{n - 1}}\left( {{{\left| \alpha  \right|}^{n - 1}} + \left| {{\alpha ^{n - 2}}} \right| + ... + 1} \right)\\\\ &\qquad\qquad\qquad= {\left| z \right|^{n - 1}} \cdot n \;\;\;\;\;\;\;\;  \left( {\left| \alpha  \right| = 1} \right)\\\\ &\Rightarrow \,\,\,\left| {{\mathop{\rm Im}\nolimits} ({z^n})} \right| \le n\left| {{\mathop{\rm Im}\nolimits} (z)} \right|\,\,\,{\left| z \right|^{n - 1}}\end{align}\]

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