# $\sqrt{-1}$ ?

If $\sqrt{-1}$ can be written as,

$\sqrt{-1} = \sqrt{1-2} = (1-2)^\dfrac{1}{2}$.

Now ,

$(1-2)^\dfrac{1}{2} = 1-\dfrac{1}{2}*(2)- \dfrac{1*1}{2*2*2}*(2^2)-\dfrac{1*1*3}{2*2*2*3!}*(2^3)-\cdots$

The above series will go upto infinite. So is this true? Is the value of $\sqrt{-1}$ is possible. Note by Akash Shukla
4 years, 4 months ago

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The convergence of the binomial expansion

$(1+x)^n = 1+ nx + \dfrac{n(n-1)}{2!} x^2 + \dfrac{n(n-1)(n-2)}{3!} x^3 + \cdots$

is only true for $-1\leq x<1$.

- 4 years, 4 months ago

Thank you very much. Now I got this. But the above series is divergence , then can we apply this?

- 4 years, 4 months ago

No. The binomial theorem is valid only when the power is an integer.

- 4 years, 4 months ago

Then what we use for expansion $\sqrt{x+1}$ or such others quantities.

- 4 years, 4 months ago