Imaginary i to the ability of i is an actual quantity – the proper manner – Thoughts Your Selections

To resolve 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 simply have a brand new toy. They experiment and play with it. So ultimately somebody questioned what’s iⁱ equal to? Extremely it’s a actual quantity!

iⁱ
= e^-π/2
≈ 0.208

There may be additionally a multi-valued consequence the place we add any a number of of 2π to the angle.

iⁱ
= e^{-π/2 + 2πokay}
okay an integer

That is an unbelievable consequence. However why is it true? I used to be searching English language movies on YouTube for a proof, and to my shock each single main video is incorrect within the sense there may be 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 incorrect after which current a extra cautious evaluation.

As standard, watch the video for an answer.

Calculating i to the ability i the proper manner

Or hold studying.
.
.

“All shall be effectively in the event you use your thoughts on your selections, and thoughts solely your selections.” Since 2007, I’ve devoted my life to sharing the enjoyment of recreation concept and arithmetic. MindYourDecisions now has over 1,000 free articles with no adverts due to neighborhood help! Assist out and get early entry to posts with a pledge on Patreon.

.
.

.
.
.
.
M
I
N
D
.
Y
O
U
R
.
D
E
C
I
S
I
O
N
S
.
.
.
.
Imaginary i to the ability of i is an actual quantity – the proper manner

(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 right them, thanks).

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

Listed here are polar types of some frequent numbers.

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

Taking okay = 0 in equations 1 and three and okay = 1 in equation 2 we now have:

i = e^πi/2
1 = e^2πi
-1 = e^πi

“Proof” 1

i = e^{i(π/2 + 2πokay)}

Increase each side to the ability i

iⁱ = (e^{i(π/2 + 2πokay)})ⁱ

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

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

For okay = 0 we get the principal worth:

iⁱ = e^-π/2

So what’s incorrect with this supposed proof? The exponent rule (a^b)^c = a^bc is just not all the time true: it could be said with some restriction like a is a constructive actual quantity and a and b are actual numbers. It isn’t true on the whole for advanced numbers. Right here is how issues can go incorrect.

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 now have proven 1 = -1 which is clearly incorrect!

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

“Proof” 2

The definition of advanced exponentiation is:

z^w = e^{w ln z}

So we now have:

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 now have discovered the right reply, however sadly the tactic is just not fully right. The step ln e^x = x required x to be an actual quantity. If not, we might find yourself with incorrect outcomes. For instance:

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

That is clearly incorrect, so we can not use this step. So how precisely are we purported to calculate i to the ability of i?

Complicated logarithm

We begin with the ability collection:

e^z = 1 + z/1! + z²/2! + z³/3! + …

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

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

So we will use this to outline advanced exponentiation as:

z^w = e^{w ln z}

However we now have truly simply created a brand new drawback: we have to outline the advanced logarithm. Recall the graph of a fancy quantity:

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

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

Suppose we now have an equation:

z = e^w

The worth w shall be equal to ln z. We are able to write z in its different polar kind and likewise 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 advanced logarithm. Let’s write Ln for the abnormal actual quantity pure logarithm perform, wherein the traditional exponent guidelines do apply. Since u is an actual quantity we will “deliver it down” and we now have:

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

Since we now have:

ln z = w = u + vi

We are able to outline the advanced 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 now have:

ln z = Ln|z| + i Arg z

Cautious evaluation

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

z^w = e^{w ln z}

So we now have:

iⁱ
= e^{i ln i}

We now calculate ln i.

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

Substituting into the exponentiation equation:

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

That is the multi-valued consequence, and all values are actual numbers. The principal worth shall be when okay = 0 and it equal to:

iⁱ
= e^-π/2
&approx; 0.208

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

Particular thanks this month to:

Daniel Lewis
Kyle
Lee Redden
Mike Robertson

Because of 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

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

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

MY BOOKS

If you are going to buy by means of these hyperlinks, I could also be compensated for purchases made on Amazon. As an Amazon Affiliate I earn from qualifying purchases. This doesn’t have an effect on the value you pay.

Guide scores are from January 2023.

(US and worldwide hyperlinks)
https://mindyourdecisions.com/weblog/my-books

Thoughts Your Selections is a compilation of 5 books:

(1) The Pleasure of Recreation Principle: An Introduction to Strategic Considering
(2) 40 Paradoxes in Logic, Likelihood, and Recreation Principle
(3) The Irrationality Phantasm: How To Make Good Selections And Overcome Bias
(4) The Greatest Psychological Math Tips
(5) Multiply Numbers By Drawing Strains

The Pleasure of Recreation Principle reveals how you should utilize math to out-think your competitors. (rated 4.3/5 stars on 290 evaluations)

40 Paradoxes in Logic, Likelihood, and Recreation Principle comprises thought-provoking and counter-intuitive outcomes. (rated 4.2/5 stars on 54 evaluations)

The Irrationality Phantasm: How To Make Good Selections And Overcome Bias is a handbook that explains the various methods we’re biased about decision-making and affords strategies to make sensible selections. (rated 4.1/5 stars on 33 evaluations)

The Greatest Psychological Math Tips teaches how one can appear like a math genius by fixing issues in your head (rated 4.3/5 stars on 116 evaluations)

Multiply Numbers By Drawing Strains This e book is a reference information for my video that has over 1 million views on a geometrical technique to multiply numbers. (rated 4.4/5 stars on 37 evaluations)

Thoughts Your Puzzles is a group of the three “Math Puzzles” books, volumes 1, 2, and three. The puzzles matters embody the mathematical topics together with geometry, likelihood, logic, and recreation concept.

Math Puzzles Quantity 1 options traditional mind teasers and riddles with full options for issues in counting, geometry, likelihood, and recreation concept. Quantity 1 is rated 4.4/5 stars on 112 evaluations.

Math Puzzles Quantity 2 is a sequel e book with extra nice issues. (rated 4.2/5 stars on 33 evaluations)

Math Puzzles Quantity 3 is the third within the collection. (rated 4.2/5 stars on 29 evaluations)

KINDLE UNLIMITED

Lecturers and college students around the globe usually e-mail me concerning the books. Since schooling can have such a huge effect, I attempt to make the ebooks accessible as extensively as doable at as low a worth as doable.

Presently you may learn most of my ebooks by means of Amazon’s “Kindle Limitless” program. Included within the subscription you’ll get entry to tens of millions of ebooks. You do not want a Kindle machine: you may set up the Kindle app on any smartphone/pill/laptop/and so forth. I’ve compiled hyperlinks to applications in some international locations under. Please verify your native Amazon web site for availability and program phrases.

US, checklist of my books (US)
UK, checklist of my books (UK)
Canada, e book outcomes (CA)
Germany, checklist of my books (DE)
France, checklist of my books (FR)
India, checklist of my books (IN)
Australia, e book outcomes (AU)
Italy, checklist of my books (IT)
Spain, checklist of my books (ES)
Japan, checklist of my books (JP)
Brazil, e book outcomes (BR)
Mexico, e book outcomes (MX)

MERCHANDISE

Seize a mug, tshirt, and extra on the official web site for merchandise: Thoughts Your Selections at Teespring.