Complex Numbers Set 6

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Example- 21

Prove that \(\begin{align}\left| {\frac{{z - {z_1}}}{{z - {z_2}}}} \right| = k\end{align}\) represents a circle if \(k \ne 1\)  and a line if k = 1.

Solution: For k = 1, the proof is trivial. The given equation reduces to

\[|z - {z_1}|\, = \,\,|z - {z_2}|\]

which implies that z lies on the perpendicular bisector of the line segment joining Z1 and Z2

Let us now consider the case when \(k \ne 1\).

We need to find the locus of a point P which moves such that

\[\frac{{AP}}{{PB}} = K\]

Note that one such point also lies on the line segment AB itself which divides AB internally in the ratio k : 1. Also, another such point lies on the extended line of the line segment AB, which will externally divide  AB in the ratio  k : 1.  If k > 1, this external point will lie on the ‘right’ side of B (in the figure above), and if k < 1, this point will lie to the ‘left’ of A.

Assuming k > 1, we let I be the point of internal division and E the point of external division of the segment AB in the ratio k : 1.

Thus, \(\begin{align}\frac{{AI}}{{IB}} = \frac{{AE}}{{EB}} = k.\end{align}\)

I is given by \(\begin{align}\frac{{k{z_2} + {z_1}}}{{k + 1}}\end{align}\)  while E is given by \(\begin{align}\frac{{k{z_2} - {z_1}}}{{k - 1}}\end{align}\)

Now, we need the points z which satisfy

 \[\begin{align}&|z - {z_1}|\,\, = \,\,k|z - {z_2}|\\\\&|z{{|}^{2}}+|{{z}_{1}}{{|}^{2}}-z{{\bar{z}}_{1}}-\bar{z}{{z}_{1}}={{k}^{2}}|z{{|}^{2}}+{{k}^{2}}|{{z}_{2}}{{|}^{2}}-{{k}^{2}}z{{\bar{z}}_{2}}-{{k}^{2}}\bar{z}{{z}_{2}}\\\\&\Rightarrow({{k}^{2}}-1)|z{{|}^{2}}-z({{k}^{2}}{{\bar{z}}_{2}}-{{\bar{z}}_{1}})-\bar{z}({{k}^{2}}{{z}_{2}}-{{z}_{1}})+{{k}^{2}}|{{z}_{2}}{{|}^{2}}-|{{z}_{1}}{{|}^{2}}=0\\\\&\Rightarrow |z{{|}^{2}}-z\frac{({{k}^{2}}{{{\bar{z}}}_{2}}-{{{\bar{z}}}_{1}})}{{{k}^{2}}-1}-\bar{z}\frac{({{k}^{2}}{{z}_{2}}-{{z}_{1}})}{{{k}^{2}}-1}+\frac{{{k}^{2}}|{{z}_{2}}{{|}^{2}}-|{{z}_{1}}{{|}^{2}}}{{{k}^{2}}-1}=0\end{align}\] 

 

\[\begin{align}&\Rightarrow {{\left| z-\frac{{{k}^{2}}{{z}_{2}}-{{z}_{1}}}{{{k}^{2}}-1} \right|}^{2}}=\frac{{{k}^{2}}(|{{z}_{1}}{{|}^{2}}+|{{z}_{2}}{{|}^{2}}-{{z}_{1}}{{{\bar{z}}}_{2}}-{{{\bar{z}}}_{1}}{{z}_{2}})}{{{({{k}^{2}}-1)}^{2}}} =\frac{{{k}^{2}}|{{z}_{1}}-{{z}_{2}}{{|}^{2}}}{{{({{k}^{2}}-1)}^{2}}}\\&\Rightarrow \left| z-\frac{{{k}^{2}}{{z}_{2}}-{{z}_{1}}}{{{k}^{2}}-1} \right|=\frac{k|{{z}_{1}}-{{z}_{2}}|}{{{k}^{2}}-1}\end{align}\]

This is the equation of a circle with centre at \(\begin{align}&\frac{{{k^2}{z_2} - {z_1}}}{{{k^2} - 1}}\end{align}\) and \(\begin{align}&\frac{{k|{z_1} - {z_2}|}}{{{k^2} - 1}}\end{align}\) radius. Note that the midpoint of I and E is

\[\frac{1}{2}\left( \frac{k{{z}_{2}}+{{z}_{1}}}{k+1}+\frac{k{{z}_{2}}-{{z}_{1}}}{k-1} \right)\]

\[ = \frac{{{k^2}{z_2} - {z_1}}}{{{k^2} - 1}}\]

Thus, the center of this circle is actually the mid-point of IE and the radius is \(\begin{align}\frac{k}{{{k^2} - 1}}\end{align}\)  times the original line-segment AB.

Example-  22

Find the complex numbers which simultaneously satisfy

\(\begin{align}&\left| {\frac{{z - 12}}{{z - 8i}}} \right| = \frac{5}{3}\end{align}\) and \(\begin{align}&\left| {\frac{{z - 4}}{{z - 8}}} \right| = 1\end{align}\)

Solution:  Readers who’ve followed the previous example will realise that the first equation is that of a circle while the second is that of a straight line which is the perpendicular bisector of 4 and 8, i.e., from the second equation, we know that z is of the form z = 6 + xi.

Using this in the first equation, we get

\[\left| {\frac{{ - 6 + xi}}{{6 + (x - 8)i}}} \right| = \frac{5}{3}\]

  \[\begin{array}{l} \Rightarrow  9({6^2} + {x^2}) = 25\left( {{6^2} + {{(x - 8)}^2}} \right) = 25(100 + {x^2} - 16x)\\ \Rightarrow  16{x^2} - 400x + 2176 = 0\\ \Rightarrow  {x^2} - 25x + 136 = 0\\ \Rightarrow  (x - 8)(x - 17) = 0\\ \Rightarrow  x = 8\,\,\,\,\,{\rm{or}}\,\,\,\,17\end{array}\]

Thus, \(z = 6 + 8i\,\,\,{\rm{or}}\,\,\,6 + 17i\)

Example-  23

If the points \({z_1},{z_2},.........{z_n}\) all lie on the same side of a line passing through the origin, show that the points \(\begin{align}\frac{1}{{{z_1}}},\frac{1}{{{z_2}}},......,\frac{1}{{{z_n}}}\end{align}\)  also lie on the same side of another line passing through the origin.

Solution: Let the given line be

\[az + \bar a\,\bar z = 0\]

All the points \({z_i}\) lie on the same side of this line, i.e.,

\[a{{z}_{i}}+\bar{a}\,{{\bar{z}}_{i}}>0\ \forall \;i \; or \; <0 \forall \;i\]

Dividing by \(|{z_i}{|^2} = {z_i}{\bar z_i}\)  on both sides of the inequality,   we obtain

\(\begin{align}\frac{a}{{{{\bar{z}}}_{i}}}+\frac{{\bar{a}}}{{{z}_{i}}}>0\end{align}\)  vi  or    <0 vi

\(\begin{align}\Rightarrow  \bar{a}\left( \frac{1}{{{z}_{i}}} \right)+a\overline{\left( \frac{1}{{{z}_{i}}} \right)}>0\end{align}\)  vi or <0  vi

All the \({z_i}'s\) lie on the same side of the line \(\bar az + a\bar z = 0\)

 Example- 24

Evaluate \(\begin{align}\sum\limits_{p = 1}^{32} {(3p + 2)} {\left( {\sum\limits_{q = 1}^{10} {\left( {\sin \frac{{2q\pi }}{{11}} - i\cos \frac{{2q\pi }}{{11}}} \right)} } \right)^p}\end{align}\)

Solution: Although the expression is enormous, the alert reader will quickly realise that this expression can be expressed in terms of the eleventh roots of unity.

\[\begin{align}{}\left( {\sin \frac{{2q\pi }}{{11}} - i\cos \frac{{2q\pi }}{{11}}} \right) =  - i\left( {\cos \frac{{2q\pi }}{{11}} + i\sin \frac{{2q\pi }}{{11}}} \right)  \end{align}\]

 \[ =  - i{\alpha ^q}{\rm{\; where\; }}\alpha  = \cos \frac{{2\pi }}{{11}} + i\sin \frac{{2\pi }}{{11}}= {e^{i2\pi /11}}\]

Thus,

  \[\begin{align}&\sum\limits_{q = 1}^{10} {\left( {\sin \frac{{2q\pi }}{{11}} - i\cos \frac{{2q\pi }}{{11}}} \right)}  = \sum\limits_{q = 1}^{10} {( - i{\alpha ^q})} \\ &\qquad\qquad\qquad\qquad\qquad\quad\;\;\;=  - i\sum\limits_{q - 1}^{10} {{\alpha ^q}} \\ &\qquad\qquad\qquad\qquad\qquad\quad\;\;\;=  - i \times  - 1 \left( {1 + \alpha  + ...{\alpha ^{10}} = 0} \right)\\ &\qquad\qquad\qquad\qquad\qquad\quad\;\;\;= i\end{align}\]

The given expression reduces to

\[\begin{array}{l}S = \sum\limits_{p = 1}^{32} {(3p + 2){i^p}} \\ = 3\sum\limits_{P = 1}^{32} {p{i^P} + 2\sum\limits_{P = 1}^{32} {{i^P}} } \end{array}\]

\(\sum\limits_{P = 1}^{32} {{i^P}} \) is obviously 0. Therefore

\[\begin{array}{l}S = 3\,[(i - 2 - 3i + 4) + (5i - 6 - 7i + 8) + ... - 31\,i + 32]\\\,\,\,\, = 3\,[( - 2i + 2) + ( - 2i + 2) + ...( - 2\,i + 2]\\\,\,\, = 3 \times 8 \times 2(1 - i)\\\,\, = 48\,(1 - i)\end{array}\]

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