complex roots multi (1 Viewer)

mr.habibbi · Aug 24, 2021

Hi All,

Answer says C, but D also works right?

$\sqrt{i^3}=\sqrt{-1}=\pm{\frac{1}{\sqrt{2}}(1-i)=\pm{\frac{1}{\sqrt{2}}(-1+i) \quad \theta=\frac{3\pi}{4}, \frac{-\pi}{4}$

Screen Shot 2021-08-22 at 7.08.51 pm.png

Life'sHard · Aug 24, 2021

i^3 is -i

mr.habibbi · Aug 24, 2021

Life'sHard said:
i^3 is -i

typo

A1La5 · Aug 24, 2021

I'm not familiar with the method you've use but here's how I would do it:

Fairly sure D would be the ~~conjugate~~ opposite of C. As the imaginary parts have opposite signs due to sine being an odd function.
Edit: Forgive me, I forgot cosine was negative in the second quadrant. This is what happens when you do too much English. -.-

CM_Tutor · Aug 24, 2021

Actually, D is -1 x C.

C has negative real part and a positive imaginary part whereas D has a positive real part and a negative imaginary part.

mr.habibbi · Aug 24, 2021

CM_Tutor said:
Actually, D is -1 x C.

C has negative real part and a positive imaginary part whereas D has a positive real part and a negative imaginary part.

if you find the solution to the complex root, you get a plus minus (±) result, so you can take out or add a minus to (1-i) to get (-1+i) and still get the same result? hence C and D should work right

Wizjaro · Aug 25, 2021

Yea I think both C and D are valid answers. After all square roots usually have 2 viable answers when involving complex numbers.

CM_Tutor · Aug 25, 2021

No, it is not. $e^{i\theta}$ has a single value.

$\begin{align*} i^3 &= \left(e^{\frac{i\pi}{2}}\right)^3 \\ &= e^{\frac{3i\pi}{2}} \\ \sqrt{i^3} &= \sqrt{e^{\frac{3i\pi}{2}}} \\ &= \left(e^{\frac{3i\pi}{2}}\right)^{\frac{1}{2}} \\ &= e^{\frac{3i\pi}{4}} \\ &= \cos{\cfrac{3\pi}{4}} + i\sin{\cfrac{3\pi}{4}} \qquad \qquad \text{answer C} \\ &= -\cfrac{1}{\sqrt{2}} + \cfrac{i}{\sqrt{2}} \\ \text{So} \qquad \sqrt{i^3} &= \cfrac{1}{\sqrt{2}}(-1 + i) \end{align*}$

$\begin{align*} \text{Now, answer D is} \qquad e^{-\frac{i\pi}{4}} &= \cfrac{1}{e^{\frac{i\pi}{4}}} \\ &= \cfrac{e^{\frac{i\pi}{4}}}{\left(e^{\frac{i\pi}{4}}\right)^2} \\ &= \cfrac{e^{\frac{i\pi}{4}}}{e^{\frac{i\pi}{2}}} \\ &= \cfrac{e^{\frac{i\pi}{4}}}{i} \\ &= \cfrac{ie^{\frac{i\pi}{4}}}{i^2} \\ &= \cfrac{ie^{\frac{i\pi}{4}}}{-1} \\ &= -ie^{\frac{i\pi}{4}} \\ &= -i\left(\cos{\cfrac{\pi}{4}} + i\sin{\cfrac{\pi}{4}}\right) \\ &= -\cfrac{i}{\sqrt{2}}(1+i) \\ &= -\cfrac{1}{\sqrt{2}}\left(i + i^2\right) \\ &= -\cfrac{1}{\sqrt{2}}(i - 1) \\ &= \cfrac{1}{\sqrt{2}}(1 - i) \\ &\neq \cfrac{1}{\sqrt{2}}(-1 + i) \\ \text{Thus} \qquad e^{-\frac{i\pi}{4}} &\neq \sqrt{i^3} \end{align*}$

mr.habibbi · Aug 25, 2021

CM_Tutor said:
No, it is not. $e^{i\theta}$ has a single value.

$\begin{align*} i^3 &= \left(e^{\frac{i\pi}{2}}\right)^3 \\ &= e^{\frac{3i\pi}{2}} \\ \sqrt{i^3} &= \sqrt{e^{\frac{3i\pi}{2}}} \\ &= \left(e^{\frac{3i\pi}{2}}\right)^{\frac{1}{2}} \\ &= e^{\frac{3i\pi}{4}} \\ &= \cos{\cfrac{3\pi}{4}} + i\sin{\cfrac{3\pi}{4}} \qquad \qquad \text{answer C} \\ &= -\cfrac{1}{\sqrt{2}} + \cfrac{i}{\sqrt{2}} \\ \text{So} \qquad \sqrt{i^3} &= \cfrac{1}{\sqrt{2}}(-1 + i) \end{align*}$

$\begin{align*} \text{Now, answer D is} \qquad e^{-\frac{i\pi}{4}} &= \cfrac{1}{e^{\frac{i\pi}{4}}} \\ &= \cfrac{e^{\frac{i\pi}{4}}}{\left(e^{\frac{i\pi}{4}}\right)^2} \\ &= \cfrac{e^{\frac{i\pi}{4}}}{e^{\frac{i\pi}{2}}} \\ &= \cfrac{e^{\frac{i\pi}{4}}}{i} \\ &= \cfrac{ie^{\frac{i\pi}{4}}}{i^2} \\ &= \cfrac{ie^{\frac{i\pi}{4}}}{-1} \\ &= -ie^{\frac{i\pi}{4}} \\ &= -i\left(\cos{\cfrac{\pi}{4}} + i\sin{\cfrac{\pi}{4}}\right) \\ &= -\cfrac{i}{\sqrt{2}}(1+i) \\ &= -\cfrac{1}{\sqrt{2}}\left(i + i^2\right) \\ &= -\cfrac{1}{\sqrt{2}}(i - 1) \\ &= \cfrac{1}{\sqrt{2}}(1 - i) \\ &\neq \cfrac{1}{\sqrt{2}}(-1 + i) \\ \text{Thus} \qquad e^{-\frac{i\pi}{4}} &\neq \sqrt{i^3} \end{align*}$

so if we were asked to find the solutions to $\sqrt{-i}$ we wouldn't put the plus minus?

CM_Tutor · Aug 25, 2021

mr.habibbi said:
so if we were asked to find the solutions to $\sqrt{-i}$ we wouldn't put the plus minus?

Firstly, I was showing that the values (C) and (D) are different.

But, to answer your question, and remembering that $-i$ is just a number, how many values does $\sqrt{2}$ have? The equation $x^2 - 2 = 0$ has two solutions but only one them is represented by the expression $\sqrt{2}$ .

The equation $z^2 + i = 0$ has two solutions, $z = \pm \cfrac{1}{\sqrt{2}}(1 - i)$ , but only one of them is the "positive" square root.

As for how to answer an exam question, bear in mind that context matters and that questions can be ambiguous / open to multiple interpretations.

~~And as far as this question goes, in choosing the "best" answer, it's clear that (C) is a better answer than (D).~~

_{Edited 27 August 2021 to strike above conclusion, after changing my view - for explanation, see below}

Trebla · Aug 25, 2021

I don't think the question is well written.

If we want to be more technical about it, the problem lies in how you actually define a “square root” of a complex number which is a bit beyond the syllabus.

Strictly speaking (i.e. in "pure maths" speak), a square root of a number a, is the solution to $x^2=a$ . In the real number system that we're all familiar with, the principal square root is the positive root $\sqrt{a}$ but we know that the other root is the negative of the principal square root i.e. $-\sqrt{a}$ .

In complex numbers, a similar principle applies. For example, strictly speaking the imaginary unit i is defined as just the solution to $x^2=-1$ . The principal root is $\sqrt{-1}$ which is what most textbooks cite. However, we can technically also define the imaginary unit as $-\sqrt{-1}$ as it also squares to give -1.

When you have both real and imaginary parts, you find the principal root on the basis of the principal argument. If

$z=re^{i\theta}$ where $-\pi < \theta \leq \pi$

then the principal square root $\sqrt{z}$ is defined as

$\sqrt{z}=\sqrt{r}e^{i\frac{\theta}{2}}$ where $-\frac{\pi}{2} < \frac{\theta}{2} \leq \frac{\pi}{2}$

Back to the question, the solution to $z^2=-i$ is

$z=e^{i\frac{3\pi}{4}},e^{-i\frac{\pi}{4}}$
However, the principal square root $\sqrt{-i}$ is the one where the argument is in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right]$

This suggests the answer is technically $e^{-i\frac{\pi}{4}$ .

If you pay careful attention to well worded questions, they usually ask you to solve x and y where

$(x+iy)^2=a+ib$

which avoids this issue completely to get both square roots - as the syllabus only refers to finding two square roots of a complex number.

The subtle difference being that they usually don't (and shouldn't) word it in such a way to ask you to find x and y where

$x+iy=\sqrt{a+ib}$

where technically you need to think about finding the principal square roots.

dan964 · Aug 26, 2021

mr.habibbi said:
if you find the solution to the complex root, you get a plus minus (±) result, so you can take out or add a minus to (1-i) to get (-1+i) and still get the same result? hence C and D should work right

Trebla is absolutely correct. The issue has to with symbols. The actual square root sign/symbol $\sqrt{\quad}$ by definition means generally means take the positive root (or strictly the principal root). It is only justified to take both roots if the values were originally squared.

Difference between the following:
$x^2= i^3$ has 2 solutions since when we square root both sides, we take both the positive and negative root.
$x = \sqrt{i^3}$ has 1 solution (positive root only by definition of symbol $\sqrt{\quad}$ )

$\cos \theta + i \sin \theta = e^{i\theta}$ (where $\theta$ in radians)

Solution D is technically more correct. Because is we take positive root to mean positive in the reals*:
(aka the real component > 0)
$\sqrt{-1} = \frac{1}{\sqrt{2}}(1-i) = \cos$-\frac{\pi}{4}$ + i \sin \(-\frac{\pi}{4}) = e^{i (-\frac{\pi}{4})}$

(* if the real value is 0, we take the root with a positive imaginary value)

dan964 · Aug 26, 2021

CM_Tutor said:
No, it is not. $e^{i\theta}$ has a single value.

$\begin{align*} i^3 &= \left(e^{\frac{i\pi}{2}}\right)^3 \\ &= e^{\frac{3i\pi}{2}} \\ \sqrt{i^3} &= \sqrt{e^{\frac{3i\pi}{2}}} \\ &= \left(e^{\frac{3i\pi}{2}}\right)^{\frac{1}{2}} \\ &= e^{\frac{3i\pi}{4}} \\ &= \cos{\cfrac{3\pi}{4}} + i\sin{\cfrac{3\pi}{4}} \qquad \qquad \text{answer C} \\ &= -\cfrac{1}{\sqrt{2}} + \cfrac{i}{\sqrt{2}} \\ \text{So} \qquad \sqrt{i^3} &= \cfrac{1}{\sqrt{2}}(-1 + i) \end{align*}$

$\begin{align*} \text{Now, answer D is} \qquad e^{-\frac{i\pi}{4}} &= \cfrac{1}{e^{\frac{i\pi}{4}}} \\ &= \cfrac{e^{\frac{i\pi}{4}}}{\left(e^{\frac{i\pi}{4}}\right)^2} \\ &= \cfrac{e^{\frac{i\pi}{4}}}{e^{\frac{i\pi}{2}}} \\ &= \cfrac{e^{\frac{i\pi}{4}}}{i} \\ &= \cfrac{ie^{\frac{i\pi}{4}}}{i^2} \\ &= \cfrac{ie^{\frac{i\pi}{4}}}{-1} \\ &= -ie^{\frac{i\pi}{4}} \\ &= -i\left(\cos{\cfrac{\pi}{4}} + i\sin{\cfrac{\pi}{4}}\right) \\ &= -\cfrac{i}{\sqrt{2}}(1+i) \\ &= -\cfrac{1}{\sqrt{2}}\left(i + i^2\right) \\ &= -\cfrac{1}{\sqrt{2}}(i - 1) \\ &= \cfrac{1}{\sqrt{2}}(1 - i) \\ &\neq \cfrac{1}{\sqrt{2}}(-1 + i) \\ \text{Thus} \qquad e^{-\frac{i\pi}{4}} &\neq \sqrt{i^3} \end{align*}$

After line 2 you could argue that you need to change need to convert $e^{\frac{3i\pi}{2}}$ to be in the range $(-\pi, \pi]$ aka $e^{\frac{-i \pi}{2}}$
So that when you take the squareroot, the result is in the principal argument range $(-\frac{\pi}{2}, \frac{\pi}{2}]$ to ensure that the real value of the solution > 0.

So D is justified and is technically more correct. Although I would mark C and D as both correct as it is ambiguous. Which paper was this from (as I might correct it)

idkkdi · Aug 26, 2021

dan964 said:
After line 2 you could argue that you need to change need to convert $e^{\frac{3i\pi}{2}}$ to be in the range $(-\pi, \pi]$ aka $e^{\frac{-i \pi}{2}}$
So that when you take the squareroot, the result is in the principal argument range $(-\frac{\pi}{2}, \frac{\pi}{2}]$ to ensure that the real value of the solution > 0.

So D is justified and is technically more correct. Although I would mark C and D as both correct as it is ambiguous. Which paper was this from (as I might correct it)

i think this q has occurred more than once lol. not too sure though.

CM_Tutor · Aug 27, 2021

I have read the posts from @Trebla and @dan964 carefully. I believe that we are all in agreement that:

both (C), $z = e^{\frac{3\pi i}{4}}$ , and (D), $z = e^{-\frac{\pi i}{4}}$ , are solutions of $z^2 = i^3$
(C) and (D) are different complex numbers
there is an ambiguity created by the use of the square root symbol for te same reason that $x^2 = 2$ has two solutions of which $x = \sqrt{2}$ is only one. $\sqrt{2} \approx 1.414...$ is a decimal approximation to the expression for the square root of two, while we would all recognise that $\sqrt{2} \approx -1.414...$ is not... but extending the convention to which is the "positive" square root of a complex number is a matter of definition.

To be completely honest, neither the definition that the principal square root must have a positive real part nor that it must have an argument that is an element of $\left(-\cfrac{\pi}{2},\ \cfrac{\pi}{2}\right]$ strike me as definitive on first look. However, I have come to the conclusion that both are natural consequences of the definition that does make sense to me. Consequently, I withdraw my earlier statement that (C) is the best answer and agree that Dan and Trebla are correct that (D) is the most technically correct answer.

My reasoning reaches their conclusion from a slightly different route, which is that the principal square root of a complex number in mod-arg form is the square root taken on the complex number expressed that is itself expressed with its principal argument. Thus:

$\begin{align*} \text{Since}\ \operatorname{Arg}\left(i^3\right) = -\cfrac{\pi}{2},\ \text{as}\ \operatorname{Arg}(z) \in \left(-\pi,\ \pi\right],\ i^3 &= e^{-\frac{i\pi}{2}} \\ \implies \sqrt{i^3} &= \sqrt{e^{-\frac{i\pi}{2}}} \\ &= \left(e^{-\frac{i\pi}{2}}\right)^{\frac{1}{2}} \\ &= e^{-\frac{i\pi}{4}} \end{align*}$

The flaw in my earlier reasoning was that, in expressing $i^3$ as $e^{\frac{3i\pi}{2}}$ , I did not restrict it to being expressed with its principal argument before taking the square root; rather, I only considered whether the result had a principal argument. Restricting the complex number to its principal argument before taking the square root has the consequences that the restrictions suggested by Trebla and Dan must be satisfied.

dan964 · Aug 28, 2021

CM_Tutor said:
I have read the posts from @Trebla and @dan964 carefully. I believe that we are all in agreement that:

both (C), $z = e^{\frac{3\pi i}{4}}$ , and (D), $z = e^{-\frac{\pi i}{4}}$ , are solutions of $z^2 = i^3$

(C) and (D) are different complex numbers

there is an ambiguity created by the use of the square root symbol for te same reason that $x^2 = 2$ has two solutions of which $x = \sqrt{2}$ is only one. $\sqrt{2} \approx 1.414...$ is a decimal approximation to the expression for the square root of two, while we would all recognise that $\sqrt{2} \approx -1.414...$ is not... but extending the convention to which is the "positive" square root of a complex number is a matter of definition.

To be completely honest, neither the definition that the principal square root must have a positive real part nor that it must have an argument that is an element of $\left(-\cfrac{\pi}{2},\ \cfrac{\pi}{2}\right]$ strike me as definitive on first look. However, I have come to the conclusion that both are natural consequences of the definition that does make sense to me. Consequently, I withdraw my earlier statement that (C) is the best answer and agree that Dan and Trebla are correct that (D) is the most technically correct answer.

My reasoning reaches their conclusion from a slightly different route, which is that the principal square root of a complex number in mod-arg form is the square root taken on the complex number expressed that is itself expressed with its principal argument. Thus:

$\begin{align*} \text{Since}\ \operatorname{Arg}\left(i^3\right) = -\cfrac{\pi}{2},\ \text{as}\ \operatorname{Arg}(z) \in \left(-\pi,\ \pi\right],\ i^3 &= e^{-\frac{i\pi}{2}} \\ \implies \sqrt{i^3} &= \sqrt{e^{-\frac{i\pi}{2}}} \\ &= \left(e^{-\frac{i\pi}{2}}\right)^{\frac{1}{2}} \\ &= e^{-\frac{i\pi}{4}} \end{align*}$

The flaw in my earlier reasoning was that, in expressing $i^3$ as $e^{\frac{3i\pi}{2}}$ , I did not restrict it to being expressed with its principal argument before taking the square root; rather, I only considered whether the result had a principal argument. Restricting the complex number to its principal argument before taking the square root has the consequences that the restrictions suggested by Trebla and Dan must be satisfied.

Very well put. One of the main motivations for defining the principal value Arg is to be able to write complex numbers in modulus-argument form. Hence for any complex number z,

$z=\left|z\right|e^{i\operatorname {Arg} z}.$

So by definition when using Euler's rule or De Moivre's we have to convert our results in be in that interval.

CM_Tutor · Aug 28, 2021

dan964 said:
Very well put. One of the main motivations for defining the principal value Arg is to be able to write complex numbers in modulus-argument form. Hence for any complex number z,

$z=\left|z\right|e^{i\operatorname {Arg} z}.$

So by definition when using Euler's rule or De Moivre's we have to convert our results in be in that interval.

Thank you, and I agree. Once I realised that I was not using the principal argument in expressing $\sqrt{i^3}$ , and that doing so would necessarily lead to the conditions that you and @Trebla described, I saw that you had to be correct. Thank you to you both for helping me to see where I was mistaken.

I do hope someone identifies the source of the question so that you can consider adding a note on THSC.

mr.habibbi · Aug 29, 2021

CM_Tutor said:
Thank you, and I agree. Once I realised that I was not using the principal argument in expressing $\sqrt{i^3}$ , and that doing so would necessarily lead to the conditions that you and @Trebla described, I saw that you had to be correct. Thank you to you both for helping me to see where I was mistaken.

I do hope someone identifies the source of the question so that you can consider adding a note on THSC.

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