Question

The range of f(x), where

$f(x) = {i}^{x} + {i}^{ - x} \: where \: x \: is \: natural \: number$

EVALUATE The Above

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Answer 1

$\large\underline{\sf{Solution-}}$

Given complex function is

$\rm :\longmapsto\:f(x) = {i}^{x} + {i}^{ - x}$

can be rewritten as

$\rm :\longmapsto\:f(x) = {i}^{x} + \dfrac{1}{ {i}^{x} }$

$\rm :\longmapsto\:f(x) = \dfrac{ {i}^{2x} + 1}{ {i}^{x} }$

$\rm :\longmapsto\:f(x) = \dfrac{ {( {i}^{2} )}^{x} + 1}{ {i}^{x} }$

We know,

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: {i}^{2} \: = \: - \: 1 \: \: }}}$

So, using this, we get

$\rm :\longmapsto\:f(x) = \dfrac{ {( - 1)}^{x} + 1}{ {i}^{x} }$

Now, it is given that x is a natural number. So, let assign some values to x.

➢ Pair of points of the above equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 1 & \sf 0 \\ \\ \sf 2 & \sf - 2 \\ \\ \sf 3 & \sf 0\\ \\ \sf 4 & \sf 2\\ \\ \sf 5 & \sf 0\\ \\ \sf 6 & \sf - 2\\ \\ \sf 7 & \sf 0\\ \\ \sf 8 & \sf 2 \end{array}} \\ \end{gathered}$

So, from this we concluded that

$\rm \implies\:\boxed{ \tt{ \: Range \: of \: f(x) = \{ - 2,0,2 \}}}$

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Explore more :-

Argument of a complex number

$\red{\begin{gathered}\boxed{\begin{array}{c|c} \bf Complex \: number & \bf arg(z) \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf x + iy & \sf  {tan}^{ - 1}\bigg |\dfrac{y}{x} \bigg|   \\ \\ \sf  - x + iy & \sf \pi - {tan}^{ - 1}\bigg |\dfrac{y}{x}\bigg | \\ \\ \sf  - x - iy & \sf  - \pi + {tan}^{ - 1}\bigg |\dfrac{y}{x}\bigg | \\ \\ \sf x - iy & \sf  - {tan}^{ - 1}\bigg |\dfrac{y}{x}\bigg | \end{array}} \\ \end{gathered}}$