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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
1 year, 8 months ago

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\[ \Large i^0=(e^{i \frac {\pi} {2}})^0=e^0=1\]

@Manish Dash Raghav Vaidyanathan · 1 year, 8 months ago

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

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 · 1 year, 8 months ago

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@Karthik Venkata Using Euler's formula

\[e^{ix} = \cos x + i \sin x,\]

we get

\[e^{i\frac{\pi}{2}} = \cos \frac{\pi}{2} + i \sin \frac{\pi}{2}\]

\[e^{i\frac{\pi}{2}} = 0 + i = i\] Sharky Kesa · 1 year, 8 months ago

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@Sharky Kesa Proof of Euler's Formula ? Karthik Venkata · 1 year, 8 months ago

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@Raghav Vaidyanathan Sir, is my remark valid in \( \mathbb{C} \) ? Please reply ! Karthik Venkata · 1 year, 8 months ago

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@Karthik Venkata First of all, don't call me sir. Raghav Vaidyanathan · 1 year, 8 months ago

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@Raghav Vaidyanathan Is there any other method to prove this? Sravanth Chebrolu · 1 year, 8 months ago

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@Raghav Vaidyanathan Thanks ! Karthik Venkata · 1 year, 8 months ago

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Yes, \( i^{0} = 1 \), see this !

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

Karthik Venkata · 1 year, 8 months ago

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@Karthik Venkata This is a classic fallacy; this equality only holds true if both \(a\) and \(b\) are positive reals. All we can say when dealing with complex numbers is that

\(\dfrac{\sqrt{a}}{\sqrt{b}} = \pm \sqrt{\dfrac{a}{b}}.\)

@Manish Dash So in your example we end up with

\(\dfrac{\sqrt{-2}}{\sqrt{3}} = i\dfrac{\sqrt{2}}{\sqrt{3}} = i\sqrt{\dfrac{2}{3}}\) and \(\dfrac{\sqrt{2}}{\sqrt{-3}} = \dfrac{\sqrt{2}}{i\sqrt{3}} = -\sqrt{\dfrac{2}{3}},\) and thus \(\dfrac{\sqrt{-2}}{\sqrt{3}} = - \dfrac{\sqrt{2}}{\sqrt{-3}}.\)

Also, you are correct in saying that \(\dfrac{i}{i} = 1.\) :) Brian Charlesworth · 1 year, 8 months ago

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@Brian Charlesworth Thanks a lot sir :) ! Karthik Venkata · 1 year, 8 months ago

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@Karthik Venkata So, is sqrt (-2) / sqrt (3) = sqrt (2) / sqrt (-3) ?? Manish Dash · 1 year, 8 months ago

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@Manish Dash You mean whether \( \dfrac{\sqrt{-2}}{\sqrt{3}} = \dfrac{\sqrt{2}}{\sqrt{-3}} \) ? Karthik Venkata · 1 year, 8 months ago

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@Karthik Venkata I hope if @Raghav Vaidyanathan can explain this. Manish Dash · 1 year, 8 months 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 · 1 year, 8 months 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 · 1 year, 8 months ago

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When you're searching for \(\dfrac ii\), you're really searching for some number \(X\) such that \(X\times i=i\). A quick check (hint: write \(X=a+bi\) for \(a,b\) real) shows that \(X=1\) is the only solution. Akiva Weinberger · 1 year, 8 months ago

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

@Manish Dash , you can see @Raghav Vaidyanathan 's answer for a precise solution of your note's heading. Sravanth Chebrolu · 1 year, 8 months ago

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@Sravanth Chebrolu i think 1^1/2= 1 or -1 am i right? Yash Shah · 1 year, 8 months ago

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@Yash Shah The square root function over the non-negative reals always produces a non-negative result. Thus \(1^{\frac{1}{2}} = \sqrt{1} = 1\) only. Over the complex numbers, with \(1 = e^{n*2\pi}\) we have that \(1^{\frac{1}{2}} = (e^{n*2\pi})^{\frac{1}{2}} = e^{n\pi},\) which equals \(1\) for even integers \(n\) and \(-1\) for odd integers. In this context we would call \(1\) the principal root. Brian Charlesworth · 1 year, 8 months ago

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@Brian Charlesworth i think sqrt{1} = [(+or -1)^{2}]^{0.5}. Yash Shah · 1 year, 8 months ago

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@Sravanth Chebrolu 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 · 1 year, 8 months ago

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