# Complex Numbers Set 9

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Example- 33

On each side of a quadrilateral ABCD, squares are drawn. The centers of the opposite squares are joined.

Show that PR and QS are equal in length and perpendicular to one another.

Solution:   Let the complex numbers $$2a,\;2b,\;2c,\;2d$$ represent the sides $$\overrightarrow {AB} ,\;\overrightarrow {BC} ,\;\overrightarrow {CD} \;{\rm{and}}\;\overrightarrow {DA}$$  of the quadrilateral, and let A be the origin. Note that a + b + c + d must be 0. Now, we write  in terms of a, b, c, d:

$P = a + ia,\;\;Q = 2a + b + ib,\;R = 2a + 2b + c + ic,\;\;S = 2a + 2b + 2c + d + id$

Thus,

$\overrightarrow {PR} = a\left( {1 - i} \right)\; + 2b + c(1 + i),\;\;\overrightarrow {QS} = b(1 - i) + 2c + d(1 + i)$

From these two relations and the fact that a + b + c + d = 0, it can be easily shown that $$\overrightarrow {PR} = i\,\overrightarrow {QS}$$, which immediately implies that, PR and QS are equal in length and perpendicular to each other.

Example- 34

Find the total area of the region satisfying the following constraints:

$\sin {\log _a}|z| > 0,\;\;a > 1,\;\;|z|\; < 1$

Solution:  $$\sin {\log _a}|z|\; > \;0 \Rightarrow {\log _a}|z|\; \in \;\left( {2n\pi ,\;(2n + 1)\pi } \right)$$

$\Rightarrow |z|\; \in \;\left( {{a^{2n\pi }},\;{a^{(2n + 1)\pi }}} \right)$

Since $$|z|\; < \;1,\;$$ we have

$|z|\ \in \ \left( {{a}^{-2n\pi }},\ {{a}^{-(2n-1)\pi }} \right),\ n\ \in \ \mathbb{Z},\ n\ \ge \ 1$

The region which z represents will consist of concentric rings, one corresponding to each value of n. The area  $${A_n}$$  of the nth ring is:

${A_n} = \pi \left( {{a^{ - (4n - 2)\pi }} - {a^{ - 4n\pi }}} \right)$

The required total area A is

\begin{align}&A = \sum\limits_{n = 1}^\infty {} {A_n} = \pi \left( {\sum\limits_{n = 1}^\infty {} {a^{ - (4n - 2)\pi }} - \sum\limits_{n = 1}^\infty {} {a^{ - 4n\pi }}} \right)\\ &\qquad\qquad\quad= \pi \left( {\frac{{{a^{ - 2\pi }} - {a^{ - 4\pi }}}}{{1 - {a^{ - 4\pi }}}}} \right)\\ &\qquad\qquad\quad= \pi \left( {\frac{{{a^{2\pi }} - 1}}{{{a^{4\pi }} - 1}}} \right)\end{align}

Example- 35

Let $$f(x) = \mathop {\mathop \Pi \limits^{n - 1} }\limits_{i = 0} (2\cos {2^i}x - 1),$$ where n > 1. Let $$r\in \mathbb{Z}.$$

Find \begin{align}f\left( {\frac{{2\pi r}}{{{2^n} \pm 1}}} \right).\end{align}

Solution:   The trick is to write $$2\cos {2^j}x\;{\rm{as}}\;{e^{i{2^j}x}} + {e^{ - i{2^j}x}}$$

Writing $${e^{ix}}\;{\rm{as}}\;\alpha$$, we have

\begin{align}&f(x) = \left( {\alpha + \frac{1}{\alpha } - 1} \right)\;\left( {{\alpha ^2} + \frac{1}{{{\alpha ^2}}} - 1} \right)...\left( {{\alpha ^{{2^{n - 1}}}} + \frac{1}{{{\alpha ^{{2^{n - 1}}}}}} - 1} \right)\\ &\;\;\;\;Z= \frac{{\left( {{\alpha ^2} - \alpha + 1} \right)\;\left( {{\alpha ^4} - {\alpha ^2} + 1} \right)...\left( {{\alpha ^{{2^n}}} - {\alpha ^{{2^{n - 1}}}} + 1} \right)}}{{{\alpha ^{1 + 2 + {2^2} + {{...2}^{n - 1}}}}}}\\ &\quad\;\;\;= \frac{1}{{{\alpha ^{{2^n} - 1}}}}\;\left\{ {\;\frac{{{\alpha ^3} + 1}}{{\alpha + 1}}\; \times \;\frac{{{\alpha ^6} + 1}}{{{\alpha ^2} + 1}} \times ... \times \frac{{{{\left( {{\alpha ^{{2^{n - 1}}}}} \right)}^3} + 1}}{{{\alpha ^{{2^{n - 1}}}} + 1}}\;} \right\}\\ \end{align}

Multiplying this by \begin{align}\frac{{{\alpha ^3} - 1}}{{\alpha - 1}}\end{align} , we have

\begin{align}&f(x) \times \frac{{{\alpha ^3} - 1}}{{\alpha - 1}} = \frac{1}{{{\alpha ^{{2^n} - 1}}}} \times \frac{{{{\left( {{\alpha ^{{2^n}}}} \right)}^3} - 1}}{{{\alpha ^{{2^n}}} - 1}}\\ &\Rightarrow f(x) = \frac{1}{{{\alpha ^{{2^n}}} - 1}}\; \cdot \;\frac{{\alpha - 1}}{{{\alpha ^3} - 1}}\; \cdot \;\frac{{{{\left( {{\alpha ^{{2^n}}}} \right)}^3} - 1}}{{{\alpha ^{{2^n}}} - 1}}\end{align}

We have therefore succeeded in finding a closed form expression for f(x). The rest is straightforward.

When \begin{align}x = \frac{{2\pi k}}{{{2^n} - 1}},\;\;{\alpha ^{{2^n} - 1}}\end{align} becomes $${e^{i(2\pi k)}}$$, that is 1. Thus,

${\alpha ^{{2^n}}} = \alpha \Rightarrow f\left( {\frac{{2\pi k}}{{{2^n} - 1}}} \right) = 1$

On the other hand, when \begin{align}x = \frac{{2\pi k}}{{{2^n} + 1}},\end{align} then \begin{align}{\alpha ^{{2^n} + 1}}\end{align}  is 1, i.e.,

${\alpha ^{{2^n}}} = \frac{1}{\alpha } \Rightarrow f\left( {\frac{{2\pi k}}{{{2^n} + 1}}} \right) = 1$

In both cases, the answer is 1.

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