Question

prove that (i)rats to power i =e-(4n+1)pi/2 and reduce it in a geometrical progression

Answer 1

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Answer 2

Concept

A geometrical progression is the sequence of numbers or terms which has constant ratio. For example, a ar, ar^2, ar^3,.....n terms, where r is the constant ratio term and a is the first term. Also from the Euler's equation we know that,

$e^{i\theta} = \cos\theta + i \sin\theta$

Given

The given expression is as follows

$i=e^{-(4n+1)\pi/2}$

where i is defined as $i=\sqrt{-1}$

Find

We have to calculate the value of i^i.

Solution

Since, $i=e^{-(4n+1)\pi/2}$

Therefore, calculating the i^i by taking the power i with i, we have

$i^{i}=(e^{-(4n+1)\pi/2})^{i}\\i^{i}=e^{-(4n+1)i\pi/2}\\i^{i}=\cos((4n+1)\pi/2)+i\sin((4n+1)\pi/2)$

Now taking the value of n as 0, 1, 2,.....to find the geometrical progression, we have

For n=o, $i=e^{-\pi/2}$

For n=1, $i=e^{-5\pi/2}$

For n=2, $i=e^{-9\pi/2}$

Hence the calculated geometrical progression is i=e^{-\pi/2}, i=e^{-5\pi/2}, i=e^{-5\pi/2},....

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