4

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 $\sqrt{i}+\sqrt{-i}$

Primo
  • 81
  • 3
    $\sqrt{a} \times \sqrt{b} = \sqrt{a \times b}$ doesn't hold when the terms have negative sign. – Osama Ghani Apr 25 '17 at 17:29
  • @OsamaGhani thanks for your answer. According to you both of my process wrong? – Primo Apr 25 '17 at 17:35
  • @OsamaGhani Actually, as a set equivalence $\sqrt{z_1z_2}=\sqrt{z_1}\sqrt{z_2}$. But we might need to evaluate the square roots on differing branches. – Mark Viola Apr 25 '17 at 18:07
  • What is $$\sqrt{z}$$ when $z$ is a complex number and not a nonnegative real number? – Did Apr 25 '17 at 19:06

3 Answers3

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 $\sqrt{-i}=-e^{-i\pi/4}$.

Mark Viola
  • 179,405
  • @D.MV is there any way to solve this without using DeMoiver's theorem? – Primo Apr 25 '17 at 17:45
  • 1
    @Primo Yes. $\sqrt{i}=z\implies z^2=i$ Then, setting real and imaginary parts equal yields $x^2-y^2=0$ and $2xy=1$. We can solve for $x$ and $y$ and find $x=\pm \frac{\sqrt 2}{2}$ and $y=\mp \frac{\sqrt 2}{2}$. – Mark Viola Apr 25 '17 at 18:05
0

I'm using your method, but I'm restraining myself from writing $\sqrt{z}$ for $z$ complex. Let's use use only $[i^2=-1]$ and $[x^2=y^2\iff x=\pm y]$

Let's have $a^2=i$ and $b^2=-i$, we are searching for the value of $(a+b)$.

$(a+b)^2=a^2+2ab+b^2=i+2ab-i=2ab$

$a^2b^2=(i)(-i)=1\iff ab=\pm 1$

  • If $ab=1$ then $(a+b)^2=2$ and $(a+b)=\pm\sqrt{2}$
  • If $ab=-1$ then $(a+b)^2=-2$ and $(a+b)=\pm i\sqrt{2}$

We found $4$ possible values, and this is normal, remember that in $\mathbb C$ each complex has $2$ possible roots so $\sqrt{z_1}+\sqrt{z_2}$ would have $4$ possible values (if $z_1\neq z_2$ of course).

Or simply notice that $z=a+b$ is solution of $z^4=(a+b)^4=(2ab)^2=4a^2b^2=4$ and $z^4-4=0$ has $4$ different roots.

zwim
  • 28,563
0

There is no definitive answer unless you specify which square root is meant by $\sqrt z$ when $z$ is a complex number. Nonetheless, we can limit the result to four possibilities.

Let $a=\sqrt i$, $b=\sqrt{-i}$, and $s=a+b$. If all we assume about $a$ and $b$ is that $a^2=i$ and $b^2=-i$, then

$$s^2=a^2+2ab+b^2=i+2ab-i=2ab$$

hence

$$s^4=4a^2b^2=4i(-i)=4$$

so we see that $s\in\{\sqrt2,-\sqrt2,\sqrt2i,-\sqrt2i\}$.

There are two standard conventions for specifying which square root is meant by the square root symbol. In one convention, $\Re(\sqrt z)\gt0$ if $z\not\in\mathbb{R}$, in which case $\sqrt i+\sqrt{-i}=\sqrt2$. In the other, $\Im(\sqrt z)\gt0$ if $z\not\in\mathbb{R}$, in which case $\sqrt i+\sqrt{-i}=\sqrt2i$.

Remark: On posting this, I see I've duplicated much of what zwim posted while I was thinking and composing, and even used the same $a^2=i, b^2=-i$ notation.

Barry Cipra
  • 79,832