MHB Quickly Calculate $\text i^{n}$ Values

  • Thread starter Thread starter Drain Brain
  • Start date Start date
AI Thread Summary
To quickly calculate powers of i, remember that i has a cyclical pattern every four exponents: i^0 = 1, i^1 = i, i^2 = -1, and i^3 = -i. This means that for any exponent n, you can find the equivalent power of i by determining the remainder of n when divided by 4. The result will correspond to one of the four values based on the remainder: 0, 1, 2, or 3. For example, i^587 can be simplified to i^3, which equals -i. Understanding this cycle allows for efficient calculations of i raised to any power.
Drain Brain
Messages
143
Reaction score
0
can you teach me how to quickly determine the values of the different powers of $\text i$ e.g $\text i^{587}$?

regards!
 
Mathematics news on Phys.org
You only need to remember that $i^0 = 1$, $i^1 = i$, $i^2 = -1$, $i^3 = -i$ (you really only need $i^2 = -1$, everything else is easily derivable) and that $i^4 = (i^2)^2 = (-1)^2 = 1$. That last relation means that if you are given a power of $i$, you can multiply or divide it by $i^4$, which does not change the result since $i^4$ is just one, but let's you add or subtract $4$ to/from the exponent. That means adding or subtracting 4 to the exponent makes no difference:
$$i^3 = i^7 = i^{11} = \cdots$$
$$i^3 = i^{-1} = i^{-5} = \cdots$$
Therefore the only thing that matters is the remainder of the exponent when divided by $4$:
$$i^k = \begin{cases} 1 ~ ~ \text{if} ~ k ~ \text{leaves a remainder of 0 when divided by} ~ 4 \\ i ~ ~ \text{if} ~ k ~ \text{leaves a remainder of 1 when divided by} ~ 4 \\ -1 ~ ~ \text{if} ~ k ~ \text{leaves a remainder of 2 when divided by} ~ 4 \\ -i ~ ~ \text{if} ~ k ~ \text{leaves a remainder of 3 when divided by} ~ 4 \end{cases}$$​
 
The powers of $i$ form what professional mathematicians call a 4-cycle, which means exactly what you think it means. So:

$i^0 = 1$ (this is sort of "by default")
$i^1 = i$
$i^2 = -1$
$i^3 = (i^2)i = (-1)i = -i$
$i^4 = (i^2)(i^2) = (-1)(-1) = 1$, and we have come "full cycle" and it just repeats forever more after this.

Multiplying by $i$ can be thought of as "rotating counter-clockwise by a quarter-turn" (this should make sense, since we repeat every 4 1/4-turns), and thus multiplying by -1 can be thought of as an "about face" (a "180" as skaters like to call it), which perhaps (finally!) explains the curious rule that:

negative*negative = positive.

Therefore:

$i^{587} = i^{4\ast146 + 3} = (i^{4\ast146})(i^3) = (i^4)^{146}(i^3) = (1^{146})(i^3) = (1)(i^3) = i^3 = -i$
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top