# Is the value of i^0 =1 ?

Respected members of Brilliant !

I was puzzled by this thought.

Kindly help me as soon as possible

Note by Manish Dash
5 years, 11 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, $\dfrac{\sqrt{a}}{\sqrt{b}} = \pm \sqrt{\dfrac{a}{b}}$

- 5 years, 11 months ago

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

- 5 years, 11 months ago

Thanks a lot sir :) !

- 5 years, 11 months ago

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

- 5 years, 11 months ago

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

- 5 years, 11 months ago

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

- 5 years, 11 months ago

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

- 5 years, 11 months ago

I hope if @Raghav Vaidyanathan can explain this.

- 5 years, 11 months ago

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

- 5 years, 11 months ago

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.

- 5 years, 11 months ago

Using Euler's formula

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

we get

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

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

- 5 years, 11 months ago

Proof of Euler's Formula ?

- 5 years, 11 months ago

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

- 5 years, 11 months ago

First of all, don't call me sir.

- 5 years, 11 months ago

Is there any other method to prove this?

- 5 years, 11 months ago

Thanks !

- 5 years, 11 months ago

and all others who see this note

- 5 years, 11 months ago

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.

- 5 years, 11 months ago

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

- 5 years, 11 months ago

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

- 5 years, 11 months ago

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.

- 5 years, 11 months ago

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

- 5 years, 11 months ago

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.

- 5 years, 11 months ago