Is the value of i^0 =1 ?

Respected members of Brilliant !

I was puzzled by this thought.

Whether i/i =1 or something else ? Please answer my query and if possible please support your statement.

Kindly help me as soon as possible

Thanking you in advance

Note by Manish Dash
4 years, 1 month ago

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Yes, i0=1 i^{0} = 1 , see this !

What I felt was ab=ab \dfrac{\sqrt{a}}{\sqrt{b}} = \sqrt{\dfrac{a}{b}} is valid in C \mathbb{C} . But as explained by Brian Charlesworth sir, ab=±ab \dfrac{\sqrt{a}}{\sqrt{b}} = \pm \sqrt{\dfrac{a}{b}}

Karthik Venkata - 4 years, 1 month ago

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This is a classic fallacy; this equality only holds true if both aa and bb are positive reals. All we can say when dealing with complex numbers is that

ab=±ab.\dfrac{\sqrt{a}}{\sqrt{b}} = \pm \sqrt{\dfrac{a}{b}}.

@Manish Dash So in your example we end up with

23=i23=i23\dfrac{\sqrt{-2}}{\sqrt{3}} = i\dfrac{\sqrt{2}}{\sqrt{3}} = i\sqrt{\dfrac{2}{3}} and 23=2i3=23,\dfrac{\sqrt{2}}{\sqrt{-3}} = \dfrac{\sqrt{2}}{i\sqrt{3}} = -\sqrt{\dfrac{2}{3}}, and thus 23=23.\dfrac{\sqrt{-2}}{\sqrt{3}} = - \dfrac{\sqrt{2}}{\sqrt{-3}}.

Also, you are correct in saying that ii=1.\dfrac{i}{i} = 1. :)

Brian Charlesworth - 4 years, 1 month ago

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Thanks a lot sir :) !

Karthik Venkata - 4 years, 1 month ago

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So, is sqrt (-2) / sqrt (3) = sqrt (2) / sqrt (-3) ??

Manish Dash - 4 years, 1 month ago

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You mean whether 23=23 \dfrac{\sqrt{-2}}{\sqrt{3}} = \dfrac{\sqrt{2}}{\sqrt{-3}} ?

Karthik Venkata - 4 years, 1 month ago

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@Karthik Venkata Yes, I do mean by that statement. As I do not know how to insert special symbols like sqrt, or pi in a comment so I had to specify my comment in that form

Manish Dash - 4 years, 1 month ago

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@Manish Dash Well, I think I am not enough experienced to answer your question of though, I have no idea !

Karthik Venkata - 4 years, 1 month ago

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@Karthik Venkata I hope if @Raghav Vaidyanathan can explain this.

Manish Dash - 4 years, 1 month ago

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i0=(eiπ2)0=e0=1 \Large i^0=(e^{i \frac {\pi} {2}})^0=e^0=1

@Manish Dash

Raghav Vaidyanathan - 4 years, 1 month ago

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Bro, can you provide the proof of your substitution :- i=eiπ2 i = e^{\frac{i\pi}{2}} and wouldn't it be a circular argument in doing so ? And also, how can you establish the equation (eiπ2)0=e0 (e^{i \frac {\pi} {2}})^0=e^0 .

Please do explain, I am a beginner in the study of complex numbers. I think this note is quite suitable to ask such fundamental questions. Thanks in advance ! I hope you don't mind me calling you 'bro' :P.

Karthik Venkata - 4 years, 1 month ago

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Using Euler's formula

eix=cosx+isinx,e^{ix} = \cos x + i \sin x,

we get

eiπ2=cosπ2+isinπ2e^{i\frac{\pi}{2}} = \cos \frac{\pi}{2} + i \sin \frac{\pi}{2}

eiπ2=0+i=ie^{i\frac{\pi}{2}} = 0 + i = i

Sharky Kesa - 4 years, 1 month ago

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@Sharky Kesa Proof of Euler's Formula ?

Karthik Venkata - 4 years, 1 month ago

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Sir, is my remark valid in C \mathbb{C} ? Please reply !

Karthik Venkata - 4 years, 1 month ago

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First of all, don't call me sir.

Raghav Vaidyanathan - 4 years, 1 month ago

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@Raghav Vaidyanathan Is there any other method to prove this?

Sravanth Chebrolu - 4 years, 1 month ago

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@Raghav Vaidyanathan Thanks !

Karthik Venkata - 4 years, 1 month ago

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I support @Karthik Venkata answer, as 11=11=1=1\frac{\sqrt{-1}}{\sqrt{-1}} = \sqrt{\frac{-1}{-1}}= \sqrt{1}=1

@Manish Dash , you can see @Raghav Vaidyanathan 's answer for a precise solution of your note's heading.

Sravanth Chebrolu - 4 years, 1 month ago

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sqrt(a*b) = sqrt(a) * sqrt(b) is valid only if at least one of a or b is non negative .... so you cannot say sqrt(-1) / sqrt(-1) = 1

Abhinav Raichur - 4 years, 1 month ago

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i think 1^1/2= 1 or -1 am i right?

Yash Shah - 4 years, 1 month ago

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The square root function over the non-negative reals always produces a non-negative result. Thus 112=1=11^{\frac{1}{2}} = \sqrt{1} = 1 only. Over the complex numbers, with 1=en2π1 = e^{n*2\pi} we have that 112=(en2π)12=enπ,1^{\frac{1}{2}} = (e^{n*2\pi})^{\frac{1}{2}} = e^{n\pi}, which equals 11 for even integers nn and 1-1 for odd integers. In this context we would call 11 the principal root.

Brian Charlesworth - 4 years, 1 month ago

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@Brian Charlesworth i think sqrt{1} = [(+or -1)^{2}]^{0.5}.

Yash Shah - 4 years, 1 month ago

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When you're searching for ii\dfrac ii, you're really searching for some number XX such that X×i=iX\times i=i. A quick check (hint: write X=a+biX=a+bi for a,ba,b real) shows that X=1X=1 is the only solution.

Akiva Weinberger - 4 years, 1 month ago

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