Imagine the imaginary!

Algebra Level 2

What is the value of

0 1 i 6

6 solutions

Jaiveer Shekhawat
Oct 28, 2014

Shall it be called cheating if I take a random value of n and plug values in the equation?

Swapnil Das - 6 years ago

It isn't "cheating", but it is an invalid proof - as you are only proving the answer for one value of n, which might be a special case (unlikely on this site but you never know) - whereas Jaiveer and others have provided a proof which works no matter which n you choose.

Tony Flury - 5 years, 8 months ago

Sanjeet Raria
Oct 28, 2014

$\large \textbf {The sum of four consecutive}$ $\large \textbf {powers of iota is always 0}$

Here the sum can be shown to be a sum of four consecutive powers of $i$ . Hence it's zero.

Renah Bernat
Nov 1, 2014

It's zero. Basic rule of adding fractions is to find the common denominator (of i^(n+3) ). Doing so will leave i^3+i^2+i+1 in the numerator. But then, i^2 = -1 and i^3 = -i, thus cancelling everything out in the numerator, leaving you with the fraction, 0/i^(n+3) = 0.

How u got i^2 as -1?!

A SUGANYA - 6 years, 7 months ago

i = square root of -1

Aljun Tusi - 6 years, 7 months ago

Rohan Mullick
Nov 4, 2014

first of all take 1/(i^n) common out of d fractions.Then 1/(i^3) and 1/i cancels out and 1 and 1/(i^2) also cancels.thus ans is 0.

Lokesh Saini
Nov 1, 2014

The given expression can be written as

(1/i^(n+3))(i^3+i^2+i+1) = (1/i^(n+3))(-i-1+i+1) = 0

Alan Pang
Nov 1, 2014

use the definition I^2=-1 and use laws of indices to give 1/i^n-1/i^n 1/i-1/i^n-1/i^n 1/i=0

Imagine the imaginary!

6 solutions

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