# 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} .

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