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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
2 years, 5 months ago

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

- 2 years, 5 months ago

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

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.

- 2 years, 5 months ago

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Using Euler's formula

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

we get

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

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

- 2 years, 5 months ago

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Proof of Euler's Formula ?

- 2 years, 5 months ago

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

- 2 years, 5 months ago

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First of all, don't call me sir.

- 2 years, 5 months ago

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Is there any other method to prove this?

- 2 years, 5 months ago

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Thanks !

- 2 years, 5 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, $$\dfrac{\sqrt{a}}{\sqrt{b}} = \pm \sqrt{\dfrac{a}{b}}$$

- 2 years, 5 months ago

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

- 2 years, 5 months ago

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Thanks a lot sir :) !

- 2 years, 5 months ago

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

- 2 years, 5 months ago

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

- 2 years, 5 months ago

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I hope if @Raghav Vaidyanathan can explain this.

- 2 years, 5 months ago

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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

- 2 years, 5 months ago

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

- 2 years, 5 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.

- 2 years, 5 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.

- 2 years, 5 months ago

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

- 2 years, 5 months ago

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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.

- 2 years, 5 months ago

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

- 2 years, 5 months 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

- 2 years, 5 months ago

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and all others who see this note

- 2 years, 5 months ago

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