To unravel the equation x² = -1, mathematicians invented a brand new quantity i such that i² = -1 or i = √-1. However mathematicians are by no means content material to only have a brand new toy. They experiment and play with it. So finally somebody questioned what’s iⁱ equal to? Extremely it’s a actual quantity!

iⁱ
= e^-π/2
≈ 0.208

There’s additionally a multi-valued end result the place we add any a number of of 2π to the angle.

iⁱ
= e^{-π/2 + 2πok}
ok an integer

That is an unbelievable end result. However why is it true? I used to be shopping English language movies on YouTube for a proof, and to my shock each single main video is unsuitable within the sense there’s at the very least 1 unjustified step (even in one in every of my shorts I make a mistake). So I’ll clarify why these “proofs” are unsuitable after which current a extra cautious evaluation.

As regular, watch the video for an answer.

Calculating i to the facility i the correct method

Or preserve studying.
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“All can be properly when you use your thoughts on your selections, and thoughts solely your selections.” Since 2007, I’ve devoted my life to sharing the enjoyment of sport idea and arithmetic. MindYourDecisions now has over 1,000 free articles with no adverts due to group help! Assist out and get early entry to posts with a pledge on Patreon.

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Imaginary i to the facility of i is an actual quantity – the correct method

(Just about all posts are transcribed shortly after I make the movies for them–please let me know if there are any typos/errors and I’ll appropriate them, thanks).

A fancy quantity z = a + bi = re^{i(θ + 2πok)} may be represented graphically:

Listed below are polar types of some frequent numbers.

i = e^{i(π/2 + 2πok)}
1 = e^i(2πok)
-1 = e^{i(π + 2πok)}

Taking ok = 0 in equations 1 and three and ok = 1 in equation 2 we’ve:

i = e^πi/2
1 = e^2πi
-1 = e^πi

“Proof” 1

i = e^{i(π/2 + 2πok)}

Elevate each side to the facility i

iⁱ = (e^{i(π/2 + 2πok)})ⁱ

Apply the exponent rule (a^b)^c = a^bc.

iⁱ = e^{i2(π/2 + 2πok)}
iⁱ = e^{-(π/2 + 2πok)}

For ok = 0 we get the principal worth:

iⁱ = e^-π/2

So what’s unsuitable with this supposed proof? The exponent rule (a^b)^c = a^bc is just not at all times true: it will be said with some restriction like a is a constructive actual quantity and a and b are actual numbers. It isn’t true basically for complicated numbers. Right here is how issues can go unsuitable.

1
= 1^1/2
= (e^2πi)^1/2

Making use of the exponent rule provides:

= e^2πi(1/2)
= e^πi
= -1

However now we’ve proven 1 = -1 which is clearly unsuitable!

So we can’t correctly justify the calculation through the use of the exponent rule.

“Proof” 2

The definition of complicated exponentiation is:

z^w = e^{w ln z}

So we’ve:

iⁱ = e^{i ln i}

We all know i = e^πi/2 is the principal worth, so we substitute:

= e^{i ln (eπi/2)}

Apply the identification ln e^x = x to get:

= e^iπi/2
= e^-π/2

So as soon as once more we’ve discovered the right reply, however sadly the tactic is just not fully appropriate. The step ln e^x = x required x to be an actual quantity. If not, we may find yourself with incorrect outcomes. For instance:

1 = e^2πi
ln 1 = ln (e^2πi)
0 = ln (e^2πi)
0 = 2πi

That is clearly unsuitable, so we can’t use this step. So how precisely are we imagined to calculate i to the facility of i?

Advanced logarithm

We begin with the facility sequence:

e^z = 1 + z/1! + z²/2! + z³/3! + …

This converges for all complicated numbers, and it additionally has the property:

e^{u + v} = e^ue^v

So we will use this to outline complicated exponentiation as:

z^w = e^{w ln z}

However we’ve really simply created a brand new drawback: we have to outline the complicated logarithm. Recall the graph of a fancy quantity:

The radius r can be equal to absolutely the worth of z, denoted |z|. And let’s write arg z to imply the angle so arg z = θ + 2πok. So we will additionally write:

z = |z| e^{i arg z}

Suppose we’ve an equation:

z = e^w

The worth w can be equal to ln z. We are able to write z in its different polar kind and in addition write w = u + vi to get:

|z| e^{i arg z}
= e^{u + vi}
= e^ue^vi

Since u and v are actual numbers, this equation implies:

|z| = e^u
arg z = v

The primary equation has actual numbers on each side of the equation, so let’s apply the common logarithm we all know and like to each side. We’ll use some notation to differentiate with the complicated logarithm. Let’s write Ln for the bizarre actual quantity pure logarithm perform, through which the conventional exponent guidelines do apply. Since u is an actual quantity we will “carry it down” and we’ve:

Ln|z| = Ln(e^u)
Ln|z| = u

Since we’ve:

ln z = w = u + vi

We are able to outline the complicated pure logarithm as:

ln z = Ln|z| + i arg z

We are able to restrict to the principal worth for the angle -π < Arg z ≤ π, so we’ve:

ln z = Ln|z| + i Arg z

Cautious evaluation

So let’s lastly calculate iⁱ. We have now the definition:

z^w = e^{w ln z}

So we’ve:

iⁱ
= e^{i ln i}

We now calculate ln i.

ln i
= Ln|i| + i Arg i
= Ln 1 + i (π/2 + 2πok)
= i(π/2 + 2πok)

Substituting into the exponentiation equation:

iⁱ
= e^{i ln i}
= e^{i·i(π/2 + 2πok)}
= e^{-(π/2 + 2πok)}

That is the multi-valued end result, and all values are actual numbers. The principal worth can be when ok = 0 and it equal to:

iⁱ
= e^-π/2
&approx; 0.208

We have now derived the right reply, and we’ve accomplished a extra cautious evaluation so our steps are justified.

Particular thanks this month to:

Daniel Lewis
Kyle
Lee Redden
Mike Robertson

Due to all supporters on Patreon!

References

WolframAlpha
https://www.wolframalpha.com/enter?i=ipercent5Ei

Math StackExchange
https://math.stackexchange.com/questions/191572/prove-that-ii-is-a-real-number
https://math.stackexchange.com/questions/1347504/for-which-complex-a-b-c-does-abc-abc-hold/
https://math.stackexchange.com/questions/1495532/when-is-abc-abc-true/1495550
https://math.stackexchange.com/questions/1347504/for-which-complex-a-b-c-does-abc-abc-hold/

https://en.wikipedia.org/wiki/Principal_value

Advanced Variables and Purposes, James Ward Brown, Ruel V. Churchill, seventh version, Chapter 1 Part 8, Roots of Advanced Numbers

The complicated logarithm, exponential and energy capabilities, Professor Howard Haber
https://scipp.ucsc.edu/~haber/ph116A/clog_11.pdf

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