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

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## Comments

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

@Manish Dash

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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 \( (e^{i \frac {\pi} {2}})^0=e^0 \).

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Using Euler's formula

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

we get

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

\[e^{i\frac{\pi}{2}} = 0 + i = i\]

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

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First of all, don't call me sir.

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

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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 \(\dfrac{i}{i} = 1.\) :)

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Thanks a lot sir :) !

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

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

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@Raghav Vaidyanathan can explain this.

I hope ifLog in to reply

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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.

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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.

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

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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.

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sqrt(a*b) = sqrt(a) * sqrt(b) is valid only if at least one of

aorbis non negative .... so you cannot say sqrt(-1) / sqrt(-1) = 1Log in to reply

@Raghav Vaidyanathan @Sandeep Bhardwaj Sir, @Sanjeet Raria Sir, @Calvin Lin Sir, @Swapnil Das , @Azhaghu Roopesh M , @Karthik Venkata , @Ronak Agarwal , @Mvs Saketh , @Pranjal Jain , @Satvik Golechha , @Pi Han Goh , @Jon Haussmann Sir, @ashutosh mahapatra , @Aditya Raut , @Pratik Shastri , @Nihar Mahajan , @Krishna Ar , @Adarsh Kumar , @Mehul Arora , @PRABIR CHAUDHURI Sir, @Gogul Raman Thirunathan , @Parth Lohomi , @Rajdeep Dhingra , @Archit Boobna , @Arron Kau Sir, @Sravanth Chebrolu , @Mehul Chaturvedi , @Chew-Seong Cheong Sir,

and all others who see this note

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