Complex Numbers Set 5

Example- 17

Show that the triangles whose vertices are \({z_1},\,\,{z_2},\,\,{z_3}{\text{ and }}{Z_1},\,\,{Z_2},\,\,{Z_3}\)  are directly similar if

\[\left| {\begin{array}{*{20}{c}}{{z_1\;\;}}{{\;Z_1\;\;}}1\\{{z_2\;\;}}{{Z_2\;\;}}1\\{{z_3\;\;}}{{Z_3\;\;}}1\end{array}} \right| = 0\]

Solution:

 

Since the triangles are directly similar, the vector \({z_1} - {z_2}\)  will be a scalar multiple of \({Z_1} - {Z_2};\) the vector \({z_2} - {z_3}\)  will be the (same) scalar multiple of \({Z_2} - {Z_3}\) and so on:

\[\begin{align}&\qquad{z_1} - {z_2} = \lambda ({Z_1} - {Z_2})\\&\qquad{z_2} - {z_3} = \lambda ({Z_2} - {Z_3})\\ &\Rightarrow \,\,\,\frac{{{z_1} - {z_2}}}{{{z_2} - {z_3}}} = \frac{{{Z_1} - {Z_2}}}{{{Z_2} - {Z_3}}}\\&\Rightarrow \,\,\,{z_1}({Z_2} - {Z_3}) + {z_2}({Z_3} - {Z_1}) + {z_3}({Z_1} - {Z_2}) = 0\end{align}\\\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\Rightarrow \,\,\,\left| {\begin{gathered}{{z_1}}{{\;\;\;Z_1}}\;\;1\\{{\;\;\;z_2}}{{\;\;\;Z_2}}\;\;\;1\\{{\;\;\;z_3\;\;}}{{\;\;\;Z_3\;\;}}1\end{gathered}} \right| = 0\]

Example- 18

Show that the perpendicular distance of a point \({{z}_{0}}\)from the line \(\bar az + a\bar z + b = 0\) \((b \in \mathbb{R})\)is

\[\frac{{\left| {a{{\bar z}_0} + \bar a{z_0} + b} \right|}}{{2\left| a \right|}}\]

Solution:

Let \({z_1}\) be the foot of the perpendicular dropped from \({z_0}\) onto the given line. We need to evaluate \(\left| {{z_0} - {z_1}} \right|\)  .

Now, since \({z_1}\) lies on the given line, we have

   \[\bar a{z_1} + a{\bar z_1} + b = 0\;\;\;\;\;\; \;\;\;\;... (1)\]

Also, the complex slopes of the given line and the perpendicular must add to 0:

\[\begin{align}&\quad\qquad\frac{{ - a}}{{\bar a}} + \frac{{{z_0} - {z_1}}}{{{{\bar z}_0} - {{\bar z}_1}}} = 0\\ &\Rightarrow \,\,\, - a({{\bar z}_0} - {{\bar z}_1}) + \bar a({z_0} - {z_1}) = 0\\ &\Rightarrow \,\,\, - a{{\bar z}_0} + \bar a{z_0} + a{{\bar z}_1} - \bar a{z_1} = 0 \;\;\;\;\;\; \;\;\;\;...  (2)\end{align}\]

From (1) + (2),

\[\begin{align}&\qquad - a{{\bar z}_0} + \bar a{z_0} + 2a{{\bar z}_1} + b = 0\\ &\Rightarrow \,\,\,{{\bar z}_1} = \frac{{a{{\bar z}_0} - \bar a{z_0} - b}}{{2a}}\\ &\Rightarrow \,\,\,{{\bar z}_1} - {{\bar z}_0} = \frac{{ - a{{\bar z}_0} - \bar a{z_0} - b}}{{2a}}\\ &\Rightarrow \,\,\,\left| {{z_1} - {z_0}} \right| = \frac{{\left| {a{{\bar z}_0} + \bar a{z_0} + b} \right|}}{{2\left| a \right|}}\end{align}\]

Example- 19

Find all  \(a\in \mathbb{R}\)  if these exists one z which satisfies \(\left| z \right| = 3,\,\,\left| {z - (a(1 + i) - i)\,} \right| \le 3\)and \(\left| {z + 2a - (a + 1)i} \right| > 3\) simultaneously.

Solution:  Since \(\left| z \right| = 3,\,\,z\)  must lie on a circle of radius 3 centred at the origin. Now, the distance of z from a(1 + i) – i must not be greater than 3, i.e., z must lie inside a circle of radius 3 centred at  a(1 + i) – i. Thus, the |z| = 3 circle and the latter circle must intersect (or at least touch) in order that both the relations |z| = 3 and \(\left| {z - (a(1 + i) - 1)} \right| \le 3\)   are  satisfied. This means that the distance between the centres of the two circles must be less than the sum of the radii, i.e., 6

\[\begin{align}{} \Rightarrow \,\,\,\left| {a(1 + i) - 1} \right| \le 6\\\\ \Rightarrow \,\,\,\left| {(a - 1) + ai} \right| \le 6\\\\ \Rightarrow \,\,\,{(a - 1)^2} + {a^2} \le 36\\\\ \Rightarrow \,\,\,2{a^2} - 2a - 35 \le 0\\\\ \Rightarrow \,\,\,\frac{{1 - \sqrt {71} }}{2} \le a \le \frac{{1 + \sqrt {71} }}{2}  \;\;\;\;\; \;\;\;\;\;   \ldots (1)\end{align}\]

By an analogous argument, for the relation \(\left| {z + 2a - (a + 1)i} \right| > 3\) to be satisfied simultaneously, z must lie outside a circle of radius 3 centred at \(\left| {z + 2a - (a + 1)i} \right| > 3\). Thus, the distance between the two centres of the two circles \(\left| z \right| = 3\) and this circle must be greater than the sum of the radii.

\[\begin{align}{} \Rightarrow \,\,\,\left| {2a - (a + 1)i} \right| > 6\\\\ \Rightarrow \,\,\,5{a^2} + 2a + 1 > 36\\\\ \Rightarrow \,\,\,a < \frac{{ - 1 - 4\sqrt {11} }}{5}{\rm{ or }}a > \frac{{ - 1 + 4\sqrt {11} }}{5}     \ldots (2)\end{align}\]

The intersection of (1) and (2) gives

\[\left[ {\frac{{1 - \sqrt {71} }}{2},\,\,\frac{{ - 1 - 4\sqrt {11} }}{5}} \right) \cup \left( {\frac{{ - 1 + 4\sqrt {11} }}{5},\,\,\frac{{1 + \sqrt {71} }}{2}} \right]\]

Example- 20

Plot the fifth roots \(16( - \sqrt 3  + i)\)  of  on the plane

Solution:   We first write \(z = 16( - \sqrt 3  + i)\) in its Euler form.

\[\begin{align}&z = 16( - \sqrt 3  + i) = 32{e^{i5\,\pi /6}}\\\\&\;\;=32{{e}^{i\left( 2p\pi +\frac{5\pi }{6} \right)}},p\in \mathbb{Z}\\\\&\Rightarrow \,\,\,{{z}^{1/5}}=2{{e}^{i\left( \frac{2p\pi }{5}+\frac{\pi }{6} \right)}},\,\,\,\,p\in \mathbb{Z}\end{align}\]

To obtain the roots, we let p take on five consecutive integral values, say p = 0, 1, 2, 3, 4. The roots obtained are:

\[2{e^{i\pi /6}},2{e^{i\pi /6}}{e^{i2\pi /5}},2{e^{i\pi /6}}{e^{i4\pi /5}},2{e^{i\pi /6}}{e^{i6\pi /5}},2{e^{i\pi /6}}{e^{i8\pi /5}}\]

These five roots will lie evenly spaced out at angles of 72° between any two consecutive roots. The first root is at an angle of 30°.