# 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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