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What the value of $i^{\frac{1}{2}}$ is ?

From the definition of exponential of complex numbers, I have $$i^{\frac{1}{2}}=e^{\frac{1}{2}\log i}.$$

And since $\log i=\log 1+\arg i =\left(\dfrac{\pi}{2}+2n\pi \right) i,$ I get
$i^{\frac{1}{2}}=e^{\frac{1}{2}(\frac{\pi}{2}+2n\pi) i}=e^{(\frac{\pi}{4}+n\pi) i} =\cos(\frac{\pi}{4}+n\pi) +i\sin(\frac{\pi}{4}+n\pi)=(-1)^n\cdot \dfrac{1+i}{\sqrt 2}.$

So, I get two values $\dfrac{1+i}{\sqrt 2}, -\dfrac{1+i}{\sqrt 2}.$

But according to wolfram alpha https://www.wolframalpha.com/input?i=i%5E%7B1%2F2%7D , $i^{\frac{1}{2}}=\dfrac{1+i}{\sqrt 2}.$

Is $-\dfrac{1+i}{\sqrt 2}$ a false value ? Do I overlook something about the definition of exponential of complex nubmers ?

daㅤ
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  • Your work is correct. This can be proven, based only on wolfram alpha, because you gave two answers: A-1, and A-2. However, [1] wolfram alpha confirmed A-1 and [2] The square of the two answers must be identical, because $(-1)^2 = 1.$ Therefore, wolfram alpha has (in effect) confirmed your work. It is probably a peculiarity of wolfram alpha that with $i = e^{i\pi/2}$, that it disfavors $e^{i5\pi/4}$ in favor of $e^{i\pi/4}.$ – user2661923 Jul 02 '22 at 07:06
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    It could be that wolfram alpha has extended the Real Analysis convention that square roots are always positive to the Complex Analysis convention that is sometimes adopted that for the principal square root, the argument $(\theta)$ is always such that $\cos(\theta)$ is always non-negative. Coupled with this convention is the convention that the principal $\sqrt{-1}$ is $(i)$, rather than $(-i)$. – user2661923 Jul 02 '22 at 07:09
  • In the WolframAlpha input field, changing the braces to parentheses will give you both answers as desired. This and this are my previous work on similar questions. – ryang Jul 02 '22 at 07:43

1 Answers1

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let $i^{\frac {1} {2}}=a=\alpha+\beta i$, because we know that this form can represent all of the complex numbers. ($\alpha, \beta \geqslant0.$)

Then, $a^2=(\alpha^2-\beta^2)+2\alpha\beta i = i.$

We can easily get that $\alpha^2-\beta^2=0, 2\alpha\beta=1.$

So, $(\alpha, \beta)=\left( \dfrac {1} {\sqrt{2}}, \dfrac {1} {\sqrt{2}} \right)$.

Therefore, $i^{\frac {1} {2}}=\dfrac{1}{\sqrt{2}}+\dfrac{i}{\sqrt{2}}$.

RDK
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