• brotherbobby
In summary, the problem is that we have to find what is ##\sqrt{i} + \sqrt{-i}##. We first tried using the Euler's formula (see Relevant Equations 1) but ran into trouble. We then tried using geometry and it turned out that the two numbers add up to ##\boldsymbol{\sqrt{2}}##. However, when we tried using algebra and tried to find the real and imaginary parts of the answer, we ran into trouble. We're not sure what the problem is, but we think it has something to do with applying rules we learned when studying ##\mathbb R ## to ##\mathbb C ##.
brotherbobby
Homework Statement
##\textbf{Evaluate : }\,\boldsymbol{\sqrt{i} + \sqrt{-i}=?}##
Relevant Equations
1. ##e^{i\theta} = \cos\theta+i\sin\theta##
2. ##\cos\theta=\frac{e^{i\theta}+e^{-i\theta}}{2}##
Statement of the problem : We have to find what is ##\sqrt{i} + \sqrt{-i}##

First Attempt (Euler's Formula) : I use the Euler's formula (see Relevant Equations 1) above which yields ##i = e^{i\frac{\pi}{2}}##. Likewise ##-i = e^{i\left(-\frac{\pi}{2}\right)}##.
Now I evaluate

where, in the last step, I used Relevant
Equation no. 2 above which follows directly from Euler's Formula.

Second Attempt (geometry) : I use the fact from attempt 1 above wherein I showed that ##i = e^{i\frac{\pi}{2}}## and ##-i = e^{i\left(-\frac{\pi}{2}\right)}## to plot the two numbers into the complex plane as shown to the right. From the image it is evident that the imaginary parts of the two numbers vanish when put together. Their real parts however "add up" to yield

So far, I have been able to establish by slightly different methods that the answer to the problem ##\textbf{Evaluate : }\,\boldsymbol{\sqrt{i} + \sqrt{-i}}## is ##\boldsymbol{\sqrt{2}}##. So the answer is a real number.

But when I use the method of surds for the same problem, I run into trouble!

Third Attempt (Algebra, the method of surds) :

Doubt : Where am I mistaken? I suppose in taking the square room of the expression in the last but one step. But we are aware that in algebra when we have ##\sqrt{f(x)}##, the conditions are that (1) ##f(x)\ge 0## and (2) ##\sqrt{f(x)}\ge 0##. Acccordingly, I took the positive square roots in both cases. Of course for the second term ##(i-1)^2 = (1-i)^2##, and hence the positive square root in those cases are ##i-1## and ##1-i##. Why take one and not the other? Is this something to do with principal values?
A hint or comment would be most welcome.

My math is not formatted. I am not sure what the problem is.

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Why did you assume that $$\sqrt{(i- 1)^2} = i - 1$$ rather than the correct (in the principal branch) $$\sqrt{(i-1)^2} = \sqrt{-2i} = \sqrt{2} e^{-i\pi/4} = 1 - i?$$

topsquark, pbuk and malawi_glenn
I do not understand. Let's leave complex numbers aside for the time being.
If I were to give you ##\sqrt{(x-y)^2}##, would you say the answer is ##(x-y)## or ##(y-x)##? That is my confusion for the moment.

brotherbobby said:
But we are aware that in algebra when we have ##\sqrt{f(x)}##, the conditions are that (1) ##f(x)\ge 0## and (2) ##\sqrt{f(x)}\ge 0##.
Do you think these conditions apply when ## f(x) ## has an imaginary part? Do you think that ## w \ge 0 ## has any meaning when ## w ## has an imaginary part?

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topsquark and malawi_glenn
brotherbobby said:
Let's leave complex numbers aside for the time being.
If I were to give you ##\sqrt{(x-y)^2}##, would you say the answer is ##(x-y)## or ##(y-x)##?
If we limit x and y to ## \mathbb R ## and have ## x \ge y ## then the answer is trivial.
brotherbobby said:
That is my confusion for the moment.
No, your confusion is that you are applying "rules" you learned when studying ## \mathbb R ## to ## \mathbb C ##.

PeroK, topsquark and malawi_glenn
You simply have to re-learn what the square root symbol means in the context of non-real numbers

topsquark
pbuk said:
Do you think these conditions apply when ## f(x) ## has an imaginary part? Do you think that ## w \ge 0 ## has any meaning when ## w ## has an imaginary part?
Yes good point. Let me clarify the case for real numbers first.
The expression ##\sqrt{(x-y)^2}## is ##\vert(x-y)\vert##. Thus it is ##(x-y)## if ##x>y## and ##(y-x)## if ##y>x##.

Now for complex numbers I am still confused. If I have ##\sqrt{(x+iy)}##, how do we go about it? It would be better to write ##x+iy = re^{i\phi}##. In that case, we have ##\sqrt{re^{i\phi}}=\sqrt{r}\sqrt{e^{i\phi}}##.

How do we decide ##\sqrt{e^{i\phi}}##?

brotherbobby said:
If I have ##\sqrt{(x+iy)}##, how do we go about it?
What does your text book/lecture notes say?

brotherbobby said:
It would be better to write ##x+iy = re^{i\phi}##.
Yes that's a good start, but note that you can do that in an infinite number of ways: ## e^{i\varphi} = e^{i(\varphi + 2n\pi)} ## so you need to choose ## \varphi ## in the same way as everyone else to get the same answer. What does your text book/lecture notes say?

malawi_glenn and topsquark
Most often if ##w,z \in \mathbb{C}## then ##\sqrt{w} = z## means that ##z\cdot z = w## and ## - \pi /2\leq\text{arg}(z) \leq \pi / 2 ## but you should, as mentioned, check your textbook for this one. Different authors uses different "branches".

For instance
##\sqrt{-\text{i}} = \dfrac{1}{\sqrt{2}} - \dfrac{\text{i}}{\sqrt{2}}## because ##\left( \dfrac{1}{\sqrt{2}} - \dfrac{\text{i}}{\sqrt{2}} \right) \cdot \left( \dfrac{1}{\sqrt{2}} - \dfrac{\text{i}}{\sqrt{2}} \right) = -\text{i}##
but hey! ##\left( -\dfrac{1}{\sqrt{2}} +\dfrac{\text{i}}{\sqrt{2}} \right) \cdot \left( -\dfrac{1}{\sqrt{2}} + \dfrac{\text{i}}{\sqrt{2}} \right) = -\text{i}## too!!! But ##\text{arg}\left( -\dfrac{1}{\sqrt{2}} +\dfrac{\text{i}}{\sqrt{2}} \right) = 3\pi/4## so...

Now you check what ##\sqrt{\text{i}}## is, and your question will be resolved.

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PeroK
More generally, an interesting question is what is ##\sqrt z + \sqrt {z^*}##?

malawi_glenn
pbuk said:
What does your text book/lecture notes say?Yes that's a good start, but note that you can do that in an infinite number of ways: ## e^{i\varphi} = e^{i(\varphi + 2n\pi)} ## so you need to choose ## \varphi ## in the same way as everyone else to get the same answer. What does your text book/lecture notes say?
Sorry to come in late. Thank you for the clarification.

Actually I picked the question online and don't have a textbook for it yet. Is it ok to say that a complex number is such that, given ##z=re^{i\phi}##, the value of ##\phi## should lie between ##[0,\pi]##. Or is it ##[0,2\pi]##?

The principle branch assumes that $\arg z \in [-\pi, \pi)$. It follows that $\arg(\sqrt{z}) \in [-\pi/2, \pi/2)$.

malawi_glenn said:
But arg(−12+i2)=3π/4 so...
Can you complete your answer? Are you therefore saying that ##\sqrt{-i}\ne \left( -\dfrac{1}{\sqrt{2}} +\dfrac{\text{i}}{\sqrt{2}} \right)##?

brotherbobby said:
Can you complete your answer? Are you therefore saying that ##\sqrt{-i}\ne \left( -\dfrac{1}{\sqrt{2}} +\dfrac{\text{i}}{\sqrt{2}} \right)##?
What do you get when you square the expression on the right side?

brotherbobby said:
Can you complete your answer? Are you therefore saying that ##\sqrt{-i}\ne \left( -\dfrac{1}{\sqrt{2}} +\dfrac{\text{i}}{\sqrt{2}} \right)##?
I left some work for you to think about. Read my post very carefully

PeroK said:
More generally, an interesting question is what is ##\sqrt z + \sqrt {z^*}##?
Let me proceed. Indeed your question is more general than mine (post#1) where ##z=i\Rightarrow z^*=-i##.
Let ##z = re^{i\phi}##
\begin{equation*}
\begin{split}
\text{Let}\;\mathscr{E}&=\sqrt{z}+\sqrt{z^*}\\
& = \sqrt{r}(\sqrt{e^{i\phi}}+\sqrt{e^{-i\phi}})\\
&=\sqrt{r}(e^{i\phi/2}+e^{-i\phi/2})\\
& = \sqrt{r}\times 2\cos\phi/2\quad\text{Using Euler's formula}\\
& = \boxed{2\sqrt{r} \cos(\phi/2)}

\end{split}
\end{equation*}

In my problem, since ##z = i\Rightarrow |z|=|z^*| = 1=r## and since ##i = e^{i\phi/2}\Rightarrow \phi = \pi/2##.
Hence the expression ##\mathscr{E} = \sqrt{i}+\sqrt{-i} = 2\cos\pi/4 = \boxed{\sqrt{2}}## .

Is this correct?

Mark44 said:
What do you get when you square the expression on the right side?
Let me do it by writing. I hope it is legible.

So indeed we find ##-i##. However, the claim is that ##\sqrt{-i}\ne (-1/\sqrt 2+i/\sqrt 2)## but rather ##\sqrt{-i}\ne (1/\sqrt 2-i/\sqrt 2)##.

I think my problem is getting resolved because of the argument involved. This was @malawi_glenn 's point in post# 9 above.

To recapitulate, ##\sqrt{-i} = e^{-i\pi/4}##. The argument of the complex number ##\sqrt{-i} = -\pi/4##. We can write that ##\sqrt{-i} = 1/\sqrt{2}-i/\sqrt{2}## not only because the latter "squares" to give ##-i## but also because its argument is the same as that of ##\sqrt{-i}##, viz. ##-\pi/4##.

On the other hand, ##\sqrt{-i} \boldsymbol{\ne} -1/\sqrt{2}+i/\sqrt{2}## even though the latter squares to give ##-i##. This is because its argument is ##3\pi/4\ne -\pi/4##, as will be evident from its position on the complex plane.

I hope my reasoning is correct.

Yes it is correct. Though as some of us mentioned, there are other choices for the "branch" so you have to control it against your textbook's definition

malawi_glenn said:
Yes it is correct. Though as some of us mentioned, there are other choices for the "branch" so you have to control it against your textbook's definition
I don't follow what other choices can there be. Let's see.

We have a (complex) number like ##\sqrt{i} = e^{i\pi/4}##. The modulus of this complex number is 1 and its argument is ##\pi/4##. Where is the choice for the argument?

Or did you mean something else?

brotherbobby said:
Or did you mean something else?
malawi_glenn said:
Most often if ##w,z \in \mathbb{C}## then ##\sqrt{w} = z## means that ##z\cdot z = w## and ## - \pi /2\leq\text{arg}(z) \leq \pi / 2 ## but you should, as mentioned, check your textbook for this one. Different authors uses different "branches".
Just look at the square root of a positive real number ##x##. We define ##\sqrt{x}## to be the positive real number such that ##\sqrt{x} \cdot \sqrt{x} = x ##. Another definition would be to say that ##\sqrt{x}## is the negative real number such that ##\sqrt{x} \cdot \sqrt{x} = x ##. Thus, in our definition of what the square root of a positive real number should equal, we made a choice of "branch".

Similarly for the square root of a non-real number, we choose a "branch" - in the definition I gave we said that the square root of a non-real number should have an argument between ##- \pi/2## and ##\pi/2##.

As you saw in the example I gave you, if we just said that ##\sqrt{-\text{i}}## should be a number which when squared, equals ##-\text{i}##. There are two numbers which satifies that, nameley ##1/\sqrt{2} - \text{i}/\sqrt{2}## and ##-1/\sqrt{2} + \text{i}/\sqrt{2}##. Now, which one of these should we pick? That is why we need "another ingredient" to our definition of the square root of a non-real number,

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PeroK
brotherbobby said:
I don't follow what other choices can there be. Let's see.

We have a (complex) number like ##\sqrt{i} = e^{i\pi/4}##. The modulus of this complex number is 1 and its argument is ##\pi/4##. Where is the choice for the argument?

Or did you mean something else?

Adding $2\pi$ to the argument of a complex number does not change its real or imaginary parts, but does add $2\pi i$ to its logarithm.

This is why we need to choose an interval of width $2\pi$ in which we regard the argument as lying. The standard branch uses $[-\pi, \pi)$. The reason for this is that the jump discontinuity in $\arg z$ is then along the negative real axis, which is convenient for applications such as inverting Laplace transforms or obtaining integral representations of various special functions.

malawi_glenn

What is the square root of ##i##?

The square root of ##i## is given by the complex numbers ##\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}## and ##-\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}##. These are the two solutions to the equation ##z^2 = i## in the complex plane.

Why do we get different answers when adding square roots of ##i##?

Adding square roots of ##i## can lead to different answers due to the properties of complex numbers and the fact that square roots have multiple values. The principal square root is typically used, but the negative square root is also a valid solution, which can lead to different sums.

How do we determine which square root of ##i## to use in calculations?

The choice of which square root to use depends on the context of the problem and the conventions adopted. In many cases, the principal square root (with the positive real part) is chosen. However, both square roots are mathematically valid, and the specific application or problem may dictate which one to use.

Can the addition of square roots of ##i## be consistent?

Yes, the addition can be consistent if a standard convention is followed. By consistently using either the principal square root or the negative square root, the results will be consistent. Inconsistent results arise when switching between different square roots without a clear convention.

What is the principal square root of ##i##?

The principal square root of ##i## is ##\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}##. This is the square root with the positive real part, which is typically used as the standard or principal value in mathematical contexts.

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