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$$\begin{align} \sqrt i+\sqrt{-i} & =e^{\frac{i\pi}{2}}+e^{\frac{3i\pi}{2}}\\ &=e^{\frac{i\pi}{2}+\frac{3i\pi}{2}}\\ &=\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\\ &=i\sqrt{2} \end{align} $$

Here when asked to find the value of $\sqrt{i}+\sqrt{-i}$, he used polar form of complex numbers. And this takes $4$ values namely $2$, $-2$, $\sqrt{2}i$, $-\sqrt{2}i$. However, the steps he used looks completely fine to me, so how did he miss the other $3$ roots? What's the fault in his process?

jimjim
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madness
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  • See this answer for a discussion of the meaning of $\sqrt{z}$. In summary, under the convention that $\sqrt{z}$ is not ambiguous, you have (for example) that $\displaystyle \sqrt{i} = \sqrt{e^{i\pi/2}} = e^{i\pi/4}.$ – user2661923 Oct 02 '21 at 01:51
  • Consider, for example, the Real Analysis convention that $\sqrt{9} + \sqrt{4}$ is not $(\pm 3 \pm 2)$ but is instead $(3 + 2)$. – user2661923 Oct 02 '21 at 01:53
  • You forgot that each square root has two values , you have two square roots then that is 4 different combinations – jimjim Oct 02 '21 at 01:54
  • It probably doesn't matter, but who is "he"? – David K Oct 02 '21 at 01:54
  • @jimjim In Real Analysis, is $\sqrt{9}$ ambiguous? Isn't it reasonable to adopt a consistent convention in Complex Analysis? – user2661923 Oct 02 '21 at 01:55
  • @jimjim As pointed out in the linked question, there is a commonly-used definition of $\sqrt\cdot$ in complex analysis that defines a single-valued function. That doesn't mean you're necessarily wrong, but it means the other person isn't necessarily wrong either. It depends on which definition was given beforehand. – David K Oct 02 '21 at 01:59
  • @madness Context definitely matters here. The calculations shown are correct for one common definition of $\sqrt\cdot$ over complex numbers, incorrect for another definition. It matters greatly whether one of those definitions was stated somewhere prior to this calculation (prior in the same book, in the same paper, in the same lecture, or in the same course), and if so, which one. – David K Oct 02 '21 at 02:04
  • The angles are not unique with $+i2\pi k$. For example, what if you were to repeat the procedure with $-i=e^{-\frac{i\pi}{2}}$? – Ninad Munshi Oct 02 '21 at 03:50
  • @user2661923 : it is other way around, Reals are a subset of Complex, so what holds true for real is not necessarily holds true for complex domain, there are in complex analysis $\sqrt i$ has two very distinct values, just saying in reals something is true does not make it true for complex, in complex $\sqrt[7]{i}$ has 7 distinct values, how does that in any way relate to what is true in reals? – jimjim Oct 02 '21 at 10:55
  • @Jimjim The idea is that some Complex Analysis convention is needed in order to have the expression $\sqrt{z}$ be unambiguous. Further, whatever Complex Analysis convention is adopted, it will be expedient if this convention is consistent in its treatment of $\sqrt{z} ~: ~z \in \Bbb{R+}$ with the already adopted convention in Real Analysis for the treatment of $\sqrt{z} ~: ~z \in \Bbb{R^+}.$ ...see next comment – user2661923 Oct 02 '21 at 19:47
  • @jimjim This dovetailing of whatever Complex Analysis convention is adopted so it is consistent with the corresponding Real Analysis convention facilitates (for example) evaluating $\sqrt{9}$. Faced with the expression $\sqrt{9}$, you don't have to ask yourself whether the background is Real Analysis or Complex Analysis. So, a Complex Analysis convention was found so that (among other things) $\sqrt{9}$ would evaluate to $(3)$, rather than $(-3)$. – user2661923 Oct 02 '21 at 19:49
  • @user2661923 then what is the unambiguous value of $\sqrt[7] 1$ ? – jimjim Oct 02 '21 at 23:25
  • @jimjim First, you have to identify the value of $\theta \in (-\pi, \pi]$ such that $(1) = e^{i\theta}.$ This value is $(\theta = 0).$ Then, to identify what the common Complex Analysis convention would identify as the principal $7$-th root of $(1)$, you compute $\alpha = \frac{\theta}{7} = \frac{0}{7} = 0.$ Then, the principal $7$-th root of $(1)$ is $e^{i\alpha} = e^{i\times 0} = 1.$ – user2661923 Oct 03 '21 at 00:22
  • @user2661923 that seems useless and counter intuitive, by that logic there is no difference between any of the nth roots , that makes it ambiguous. Not sure how much complex analysis has changed in last 30 years when I did the course , but back in the day we had to be explicit ( at least at University level. Maybe in highschool it is acceptable to not bother ) – jimjim Oct 03 '21 at 02:04

1 Answers1

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There are 2 distinct values for $\sqrt {i} $,namely $p_1$ and $p_2$

$$\sqrt {i} \in \{p_1=e^{\frac{\pi i}{4}},p_2=e^{\frac{-3\pi i}{4}}\}$$

There are 2 distinct values for $\sqrt {-i} $,namely $p_3$ and $p_4$ $$\sqrt {-i} \in \{p_3=e^{\frac{-\pi i}{4}},p_4=e^{\frac{3\pi i}{4}}\}$$

so there are 4 different possible values for $\sqrt i +\sqrt{-i}$: $p_1+p_3,p_1+p_4, p_2+p_3,p_2+p_4$

jimjim
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