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- anonymous

Can somebody please explain this to me

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- anonymous

Can somebody please explain this to me

- chestercat

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- anonymous

- anonymous

This

- 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

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- anonymous

ðŸ˜’ seriously @phi

- anonymous

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

- 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

- 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)

- 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??

- phi

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

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