## anonymous one year ago Can somebody please explain this to me

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1. anonymous

@surjithayer

2. anonymous

This

3. phi

I assume you know $\sqrt{4}= \pm 2$ or $4 = (-2)^2 \text{ and } 4 = 2^2$ 2 and -2 are the 2nd root (commonly referred to as the square root) of 4

4. anonymous

😒 seriously @phi

5. anonymous

I want you to explain the concept of this formula not what does it say

6. phi

we can also write that as $4^\frac{1}{2} = \pm 2$ Let's write 4 as a complex number 4 + 0i in rectangular form in polar form $4 = 4 e^{i 2\pi} = 4( \cos(2 \pi) + \sin(2\pi) \ i)\\ = 4(1+0i)= 4$ and if we want to be complete, we should write that with an angle of 2pi n where n is any integer. $4 = 4 e^{i 2\pi\ n}$ now if we take the square root we get $4^\frac{1}{2}= \left( 4 e^{2\pi\ n} \right)^\frac{1}{2} \\ = 4^\frac{1}{2}\cdot e^{\frac{i 2\pi\ n}{2}} \\ = 2 e^{i \pi\ n}$ if n is 0 (or in general, if n is even) the exponent is a multiple of 2pi and we will get +2 if n is 1 (or odd), then exp(i pi) = cos (pi )+ i sin(pi) = -1 and we will get -2

7. phi

we can extend that idea to the "nth root" (rather than just 2) and we can do the same "true" complex numbers (not just 4, which is real)

8. anonymous

Nicee thanks @phi i have just on small thing about this app man i answer many people and get alot of medala and im still stuck at 62 from like a year ago! Why's that??

9. phi

**i answer many people and get alot of medals*** your profile shows you answer 251. Ask back after you reach 1000 or 1500