What is(are) the value(s) of $\sqrt{i}+\sqrt{-i}$?

When I am going to find out the value of $\sqrt{i}+\sqrt{-i}$, I stuck to evaluate $\sqrt{i}\times \sqrt{-i}$.

Progress: $\sqrt{i}+\sqrt{-i}=\sqrt{(\sqrt{i}+\sqrt{-i})^2}=\sqrt{2\times \sqrt{i}\times\sqrt{-i}}$.

Now $\sqrt{i}\times\sqrt{-i}=\sqrt{i\times (-i)}=\sqrt{-i\times i}=\sqrt{-1\times i^2}=\sqrt{-1\times -1}=1$
But again, $\sqrt{i}\times\sqrt{-i}=\sqrt{i}\times \sqrt{i}\sqrt{-1}=\sqrt{i}\times \sqrt{i}\times i=i\times i=i^2=-1$.

Which one is correct and what is the logic behind it?
and Finally what are the values of $\sqrt{i}+\sqrt{-i}$

Solution 1:

Note that we can write $i=e^{i(\pi/2+2k\pi)}$ for any $k\in \mathbb{Z}$. Therefore, the square root of $i$ is multivalued and can be written

$$\sqrt i=\pm e^{i\pi/4}=\pm \frac{\sqrt 2}{2}\left(1+i\right) \tag 1$$

Similarly, $-i=e^{-i(\pi/4+2k\pi)}$ and its square root is multivalued and can be written

$$\sqrt{-i}=\pm e^{-i\pi/4}=\pm \frac{\sqrt 2}{2}\left(1-i\right)\tag 2$$

Adding $(1)$ and $(2)$ from the same branch yields

$$\sqrt {i}+\sqrt{-i}=\pm \sqrt 2$$

Adding $(1)$ and $(2)$ from braches with opposing signs, we find

$$\sqrt {i}+\sqrt{-i}=\pm i \sqrt 2$$

GENERAL DISCUSSION:

To answer the more general question regarding the product $(z_1^a)(z_2^a)$, we appeal to the definition of $z^c$, where $c\in \mathbb{C}$. Then, we see that

$$z^{c}=e^{c\log(z)}$$

where $\log(z)=\log(|z|)+i\arg(z)$ is the multivalued logarithm function.

We assert, therefore, that

$$(z_1^a)(z_2^a)=(z_1z_2)^a \tag3$$

where $(3)$ is interpreted in terms of set equivalence See this answer.

This means that the product of any value of $z_1^a$ and any value of $z_2^a$ can be written as some value of $(z_1z_2)^a$. And conversely, any value of $(z_1z_2)^a$ can be expressed as the product of some value of $z_1^a$ and some value of $z_2^a$.

NOTE: It is important to understand that $(3)$ does not hold in general if $z^a$ is taken on the Principal branch of the complex logarithm (or any other designated branch since then we lose a degree of freedom).

EXAMPLE:

For $z_1=i$, $z_2=-i$, and $a=1/2$, we have

$$\sqrt{i}\sqrt{-i}=\sqrt{i(-i)}=\sqrt{1}$$

If we choose $\sqrt{i}=e^{i\pi/4}$ and $\sqrt{-i}=e^{-i\pi/4}$, then we must choose the branch of $\sqrt{z}$ for which $\sqrt{1}=1$. On the other hand, if we choose $\sqrt{i}=-e^{i\pi/4}$ and $\sqrt{-i}=e^{-i\pi/4}$, then we must choose the branch of $\sqrt{z}$ for which $\sqrt{1}=-1$.

Conversely, if we choose the branch of $\sqrt{z}$ for which $\sqrt{1}=1$, then for $\sqrt{i}=-e^{i\pi/4}$ we must have $\sqrt{-i}=-e^{-i\pi/4}$.