Examples On Arithmetic With Complex Numbers

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Example- 11

For complex numbers \(z\) and \(w,\) prove that \({\left| z \right|^2}w - {\left| w \right|^2}z = z - w\) implies that \(z = w\,\,or \;z\bar w = 1\) .

Solution: 

\[\begin{align}&{\left| z \right|^2}w - {\left| w \right|^2}z = z - w.\
\\\\&\Rightarrow \quad \frac{z}{w} = \frac{{1 + {{\left| z \right|}^2}}}{{1 + {{\left| w \right|}^2}}}\;\;\;\;\;\;\;\; ... (1) \\\\&\Rightarrow \quad \frac{z}{w}  \rm{\;is\;purely\;real.}\end{align}\]

Let,  \(z = kw\), where \(k \in \mathbb{R}\,.\) Substituting for \(z \) in (1), we get

\[\begin{align}&\frac{{kw}}{w} = \frac{{1 + {k^2}{{\left| w \right|}^2}}}{{1 + {{\left| w \right|}^2}}}\\\\ &\Rightarrow k + k{\left| w \right|^2} = 1 + {k^2}{\left| w \right|^2}\\\\&\Rightarrow k - 1 = k{\left| w \right|^2}(k - 1)\\\\&\Rightarrow(k{\left| w \right|^2} - 1)(k - 1) = 0\\\\&\Rightarrow k = 1\,\,or\;k = \frac{1}{{{{\left| w \right|}^2}}}\\\\&\Rightarrow \frac{z}{w} = 1\,\,{\rm{or }}\frac{z}{w} = \frac{1}{{{{\left| w \right|}^2}}} = \frac{1}{{w\bar w}}\\\\&\Rightarrow z = w\,\,{\rm{\;or\;}}z\bar w = 1.\end{align}\]

Example- 12

If \({z_1}\,{\rm{ and\,}}{z_2}\) are complex numbers such that \(\left| {{z_1}} \right| < 1 < \left| {{z_2}} \right|,\) prove that

\[\left| {\frac{{1 - {z_1}{{\bar z}_2}}}{{{z_1} - {z_2}}}} \right| < 1\]

Solution: We can equivalently show that  \(\left| {1 - {z_1}{{\bar z}_2}} \right| < \left| {{z_1} - {z_2}} \right|\;{\rm{ or \;}}{\left| {1 - {z_1}{{\bar z}_2}} \right|^2} < {\left| {{z_1} - {z_2}} \right|^2}\) . This is convenient because we know how to expand  \({\left| z \right|^2}( = z\bar z)\)

\[ \begin{align}&\Rightarrow {\left| {1 - {z_1}{{\bar z}_2}} \right|^2} - {\left| {{z_1} - {z_2}} \right|^2}\\\\
&= \left\{ {1 + {{\left| {{z_1}} \right|}^2}{{\left| {{z_2}} \right|}^2} - 2{\mathop{\rm Re}\nolimits} ({z_1}{{\bar z}_2})} \right\} - \left\{ {{{\left| {{z_1}} \right|}^2} + {{\left| {{z_2}} \right|}^2} - 2{\mathop{\rm Re}\nolimits} ({z_1}{{\bar z}_2})} \right\}\\\\&= 1 + {\left| {{z_1}} \right|^2}{\left| {{z_2}} \right|^2} - {\left| {{z_1}} \right|^2} - {\left| {{z_2}} \right|^2}\\\\&=  (1 - {\left| {{z_1}} \right|^2})(1 - {\left| {{z_2}} \right|^2}) < 0 \qquad (\;\rm{because} \left| {{z_1}} \right| < 1 < \left| {{z_2}} \right| )\\\\ &\Rightarrow  {\left| {1 - {z_1}{{\bar z}_2}} \right|^2} < {\left| {{z_1} - {z_2}} \right|^2}\end{align}\]

Hence, we get the desired result.

Example- 13

If \(\begin{align}\left| {z - \frac{4}{z}} \right| = 2,\end{align}\) find  \(\max \,(\left| z \right|)\,\,{\rm{and }}\min \,(\left| z \right|)\) .

 Solution:  Applying the triangle inequality on \(\left| {z - \frac{4}{z}} \right|,\) we get

\[\begin{align}&\left| {\left| z \right| - \frac{4}{{\left| z \right|}}} \right| \le \left| {z - \frac{4}{z}} \right| \le \left| z \right| + \frac{4}{{\left| z \right|}}\\\Rightarrow\quad& \left| {\left| z \right| - \frac{4}{{\left| z \right|}}} \right| \le 2 \le \left| z \right| + \frac{4}{{\left| z \right|}}\end{align}\]

The right side of this inequality is always satisfied (verify). We therefore, use the left side of this inequality:

\[\begin{align}&\qquad\;\;\left| {\left| z \right| - \frac{4}{{\left| z \right|}}} \right| \le 2\\&\Rightarrow \quad- 2 \le \left| z \right| - \frac{4}{{\left| z \right|}} \le 2\\&\Rightarrow \quad{\left| z \right|^2} + 2\left| z \right| - 4 \ge 0\,\,{\rm{\;and\;}}{\left| z \right|^2} - 2\left| z \right| - 4 \le 0\\&\Rightarrow\quad \left| z \right| \ge  - 1 + \sqrt 5 {\rm{ \;and \;}}\left| z \right| \le 1 + \sqrt 5 \\&\Rightarrow \quad\sqrt \;5\; - 1 \le \left| z \right| \le \sqrt 5  + 1\end{align}\].

These are the required maximum and minimum values.

Example- 14

If \(\left| {\frac{{1 - iz}}{{z - i}}} \right| = 1\) , show that \(z\)  is purely real.

Solution:  From the given relation, \({\left| {1 - iz} \right|^2} = {\left| {z - i} \right|^2}\)

\[\begin{align}&\Rightarrow 1 + {\left| z \right|^2} - 2{\mathop{\rm Re}\nolimits} (iz) = {\left| z \right|^2} + 1 - 2{\mathop{\rm Re}\nolimits} (i\,\bar z)\\&\Rightarrow {\mathop{\rm Re}\nolimits} (iz) = {\mathop{\rm Re}\nolimits} (i\,\bar z)\end{align}\]

If \(z = x + iy,\)  this means that

\[\begin{align}&\qquad{\mathop{\rm Re}\nolimits} (i(x + iy)) = {\mathop{\rm Re}\nolimits} (i(x - iy))\\&\Rightarrow \quad- y = y\\&\Rightarrow \quad{y = 0}\\&\Rightarrow\quad  z\;\rm{is\;purely\;real}.\end{align}\]

Example- 15

Find all non-zero complex numbers \(z\) satisfying \(\bar z = i{z^2}\) .

Solution:  We let  \(z = x + iy\)  . Using the given relation, we get

\[\begin{align}x - iy &= i{(x + iy)^2}\\& = i({x^2} - {y^2} + 2ixy)\\&=  - 2xy + i({x^2} - {y^2})\end{align}\]

Comparing the real and imaginary parts, we get

\[\begin{align}&- 2xy = x\\&\text{and}\;{x^2} - {y^2} =  - y\end{align}\]

From the first equation, we get

\[x(1 + 2y) = 0\]

\[ \Rightarrow \quad x = 0\,\,{\text{or }}y =  - 1/2\]

Now we use these two values in the second equation:

\[x = 0\,\,\, \Rightarrow \,\,\,y = 0\,\,{\rm{or \;}}1\]

  \[y =  - 1/2\,\,\, \Rightarrow \,\,\,x =  \pm \sqrt 3 /2\]

Thus,  we get the following solutions for \(x\) and \(y:\)

\[(0,\,\,0),\,\,(0,\,\,1),\,\,\left( {\frac{{\sqrt 3 }}{2},\,\,\frac{{ - 1}}{2}} \right),\,\,\left( {\frac{{ - \sqrt 3 }}{2},\,\,\frac{{ - 1}}{2}} \right)\]

Since we want a non-zero solution, we neglect the solution \((0, 0).\) The valid solutions are (in the form \(x + iy\)):

  \[i,\,\, \pm \frac{{\sqrt 3 }}{2} - \frac{1}{2}i\].

TRY YOURSELF - IV

Q. 1  Find the number of solutions to the equation \({z^2} + \bar z = 0\) .

Q. 2  If  \({z_1}\) and \({z_2}\)  are complex numbers, prove that

\(|{z_1} + {z_2}{|^2} = |{z_1}{|^2} + |{z_2}{|^2} \,\,\,\,\, \Leftrightarrow  {z_1}{\bar z_2}\)  is purely imaginary

Q. 3  Prove that if \({z_1}\;{\rm{and}}\;{z_2}\)  are two complex numbers and \(\lambda  > 0,\) then

\(|{z_1} + {z_2}{|^2} \le \left( {1 + c} \right)|{z_1}{|^2} + \left( {1 + \frac{1}{c}} \right)|{z_2}{|^2}\)

Q. 4  \({z_1}\;{\rm{and}}\;{z_2}\)  are two complex numbers such that \(\begin{align}\frac{{{z_1} - 2{z_2}}}{{2 - {z_1}\;{{\bar z}_2}}}\end{align}\) is unimodular (modulus = 1), while \({z_2}\) is not unimodular. Find  \(|{z_1}|\) .

Q. 5  Find the complex numbers \(z\) satisfying the equation \({z^2} + i\bar z = 0\) .

Q. 6  Find the maximum value of  \(|z|\)  if \(\begin{align}\left| {\;z - \frac{8}{z}\;} \right|\; = \;2\end{align}\) .