Complex Numbers Set 3

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Example-  9

Find the orthocentre of the triangle whose vertices are \({z_1},{z_2}\) and \({z_3}\)

Solution:   Let z be the required orthocentre

From the figure, it is clear that:

\[\begin{align}{}\arg \left( {\frac{{{z_1} - z}}{{{z_3} - {z_2}}}} \right) = \frac{\pi }{2}\\ \end{align}\]

\[ \Rightarrow  \frac{{{z_1} - z}}{{{z_3} - {z_2}}}\] is purely imaginary

 \[\begin{align}& \Rightarrow  \frac{{{z_1} - z}}{{{z_3} - {z_2}}} + \frac{{{{\bar z}_1} - \bar z}}{{{{\bar z}_3} - {{\bar z}_2}}} = 0\\ &\Rightarrow  \bar z - {{\bar z}_1} = \frac{{({{\bar z}_3} - {{\bar z}_2})({z_1} - z)}}{{{z_3} - {z_2}}}  \ldots (1)\end{align}\]

Similarly,  \(\begin{align}& \arg \left( {\frac{{{z_2} - z}}{{{z_1} - {z_3}}}} \right) = \frac{\pi }{2}\end{align}\)

  \[ \Rightarrow  \bar z - {\bar z_2} = \frac{{({{\bar z}_1} - {{\bar z}_3})({z_2} - z)}}{{{z_1} - {z_3}}}   \ldots (2)\]

Subtracting (2) from (1), we obtain

\[{\bar z_2} - {\bar z_1} = \frac{{({{\bar z}_3} - {{\bar z}_2})({z_1} - z)}}{{{z_3} - {z_2}}} - \frac{{({{\bar z}_1} - {{\bar z}_3})({z_2} - z)}}{{{z_1} - {z_3}}}   \ldots (3)\]

A sequence of (lengthy and tedious !) manipulations to separate z in terms of the other constants in (3) will give

\[z = \frac{{|{z_1}{|^2}({z_2} - {z_3}) + |{z_2}{|^2}({z_3} - {z_1}) + {{\left| {{z_3}} \right|}^2}({z_1} - {z_2}) + {z_1}({{\bar z}_2} - {{\bar z}_3}) + {z_2}({{\bar z}_3} - {{\bar z}_1}) + {z_3}({{\bar z}_1} - {{\bar z}_2})}}{{({{\bar z}_1}{z_2} - {z_1}{{\bar z}_2}) + ({{\bar z}_2}{z_3} - {z_2}{{\bar z}_3}) + ({{\bar z}_3}{z_1} - {z_3}{{\bar z}_1})}}\]

Example- 10

If a is the nth roots of unity given by \(\alpha  = {e^{i2\pi /n}}\) and \({z_1}\) and \({z_2}\) are any two complex numbers, show that

\[\sum\limits_{p = 0}^{n - 1} {{{\left| {{z_1} + {\alpha ^p}{z_2}} \right|}^2} = n\left( {|{z_1}{|^2} + |{z_2}{|^2}} \right)} \]

Solution:         \(|{z_1} + {\alpha ^p}{z_2}{|^2} = \left( {{z_1} + {\alpha ^p}{z_2}} \right)\left( {{{\bar z}_1} + {{\bar \alpha }^p}{{\bar z}_2}} \right)\)

Now, \(\sum\limits_{p = 0}^{n - 1} {{\alpha ^p} = 0} \) (sum of the nth roots is 0)

      \[ \Rightarrow  \sum\limits_{p = 0}^{n - 1} {{{\bar \alpha }^p} = 0} \]

Thus,

\[\begin{array}{l}\sum\limits_{p = 1}^{n - 1} {|{z_1} + {\alpha ^p}{z_2}{|^2} = n\left( {|{z_1}{|^2} + |{z_2}{|^2}} \right) + {{\bar z}_1}{z_2}\sum\limits_{p = 0}^{n - 1} {{\alpha ^p} + {z_1}{{\bar z}_2}} \sum\limits_{p = 0}^{n - 1} {{{\bar \alpha }^p}} } \\   = n\left( {|{z_1}{|^2} + |{z_2}{|^2}} \right)\end{array}\]

Example- 11

If both  \(w\;and\;{w^2}\) (non-real cube roots of unity) satisfy the equation

\[\frac{1}{{a + x}} + \frac{1}{{b + x}} + \frac{1}{{c + x}} + \frac{1}{{d + x}} = \frac{2}{x}\]

show that x = 1 also satisfies this equation.

Solution: The given equation can be re-written as (upon re-arranging into the standard form)

\[2{x^4} + (a + b + c + d){x^3} - (abc + acd + abd + bcd)x - 2abcd = 0\]

w and \({\omega ^2}\) are roots of this equation. Let a and b be the other two roots.

  \(\begin{align} \Rightarrow  \omega  + {\omega ^2} + \alpha  + \beta  =  - \frac{{(a + b + c + d)}}{2}\end{align}\)

Since \(\omega  + {\omega ^2} =  - 1,\) this reduces to

\[\alpha  + \beta  = 1 - \frac{{(a + b + c + d)}}{2}   \cdots (1)\]

Also, the coefficient of \({x^2}\) is 0

\[\begin{array}{l} \Rightarrow  \alpha \beta  + \alpha \omega  + \alpha {\omega ^2} + \beta \omega  + \beta {\omega ^2} + {\omega ^3} = 0\\ \Rightarrow  \alpha \beta  - \alpha  - \beta  + 1 = 0\\ \Rightarrow  (\alpha  - 1)(\beta  - 1) = 0\\ \Rightarrow  \alpha  = 1 {\rm{or}} \beta  = 1\end{array}\]

If \(\alpha  = 1\) , b can be determined from (1) and vice-versa

Thus, x = 1 is a root of the given equation

Example- 12

If \(|{a_n}| < 2\)  for \(1 \le n \le k,\) show that the equation

\[1 + {a_1}z + {a_2}{z^2} + ...... + {a_k}{z^k} = 0\]

cannot be satisfied by any z whose modulus is less than \(\frac{1}{3}\)

Solution:     Assume that \(|z|\, < \frac{1}{3}.\) We will use this assumption to arrive at a contradiction, proving the assumption wrong.

\[\begin{align}&p = {a_1}z + {a_2}{z^2} + .... + {a_k}{z^k}\\\\\Rightarrow\quad &|p|\, = |{a_1}z + {a_2}{z^2} + .... + {a_k}{z^k}|\\\\ &\le \,|{a_1}z| + |{a_2}{z^2}| + ..... + |{a_k}{z^k}|{\;\;\;\;\;\;\;\;\rm{(Triangle \; inequality)}}\\\\&< 2\left( {|z| + |z{|^2} + ...... + |z{|^k}} \right)(∵|{a_n}| < 2\; \forall n)\\\\ &< 2\left( {\frac{1}{3} + \frac{1}{{{3^2}}} + ...... + \frac{1}{{{3^k}}}} \right)\\\\&=2 \times \frac{1}{3}\left( {\frac{{1 - \frac{1}{{{3^k}}}}}{{1 - \frac{1}{3}}}} \right)\\\\&=1 - \frac{1}{{{3^k}}}\\\\&< 1\end{align}\]

Thus, \(|p| < 1\,\,{\rm{or}}\,\, - 1 < p < 1\)

       \(\Rightarrow \)        1 + p can never be 0, or the given equation can never

be satisfied if \(\begin{align}|z|\, < \frac{1}{3}\end{align}\) .

\(\Rightarrow \)  Our assumption\(\begin{align}|z|\, < \frac{1}{3}\end{align}\)  is wrong. Thus,

\[\begin{align}|z|\, {|\!\!\!\!\!<} \frac{1}{3}\end{align}\]

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