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To Do

Given that $\;\displaystyle w_1 \;=\; \left(2 + \sqrt{\sqrt{2}}\sqrt{2 + \sqrt{2}}\right) - i\left(1 + \sqrt{\sqrt{2}}\sqrt{2 - \sqrt{2}}\right)$.
1. Derive the two square roots of $w_1$.
2. Illustrate the general method of deriving the square roots of such a messy complex number as $w_1.$

Context

In "An Introduction to Complex Function Theory", 1991, by Bruce Palka,
problem 4.14.(iii), p26 specifies : find all roots of
$\;z^4 + (-4+2i)z^2 - 1 = 0.$

Preliminary to this problem, it is established that :

(a) Arg($z$) is the unique angle $\;\alpha \in (-\pi,\pi]\;$ such that $\;z = |z|\left[\cos(\alpha) + i\sin(\alpha)\right].$

(b) Taking $\;\beta = (\alpha/2), \;\sqrt{z} \;=\; \pm \sqrt{|z|}\left[\cos(\beta) + i\sin(\beta)\right].$

(c) $\displaystyle \cos(\beta) \;=\; \sqrt{\frac{1 + \cos(\alpha)}{2}}, \;\;\sin(\beta) \;=\; \sqrt{\frac{1 - \cos(\alpha)}{2}}.$

(d) $\;az^2 + bz + c = 0\;$ will have roots $\displaystyle\;\frac{1}{2a}\left(-b \pm \sqrt{b^2 - 4ac}\right).$

My Attack Of Problem (iii)

My first approach was :
1. let $\;w = z^2,\;$
2. interpret problem (iii) as a quadratic equation in $w$.
3. use the preliminary concepts to derive the two solutions $w_1$ and $w_2.$
4. take the two square roots of both $w_1$ and $w_2,\;$ to derive the 4 roots $\;z_1, z_2, z_3, z_4.$

One of the roots to problem (iii) interpreted as a quadratic equation, $w_1,$ is as identified in the To Do section at the start of this query.

However, after identifying $w_1$ and assigning $\;\alpha \;=\; \text{Arg}(w_1), \;$ I was unable to compute $\;\cos(\alpha)\;$ or $\;\sin(\alpha).\;$ Since Palka's preliminary concepts didn't seem to help here, I temporarily abandoned this approach.

My second approach, which succeeded, and was probably the intended approach, was :
1. factor $\;z^4 + (-4+2i)z^2 - 1 \;=\; (z^2 + 2z + i) \times (z^2 - 2z + i).$
2. solve each of the two resulting quadratic equations.

Solving both of these quadratic equations, I generated four roots, one of which was
$\displaystyle z_2 \;=\; \left(-1 - \frac{1}{2}\sqrt{\sqrt{2}}\sqrt{2 + \sqrt{2}}\right) \;+\; i \, \left(\frac{1}{2}\sqrt{\sqrt{2}}\sqrt{2 - \sqrt{2}}\right).$

After manually verifying that $z_2$ did satisfy problem (iii), I noticed that $\;(z_2)^2 = w_1,\;$ which provided a separate verification of $z_2.$

However, I feel that I should not have had to abandon the first approach. I think that there should be a way of $\underline{\text{deriving}}$ that $z_2$ is one of the square roots of $w_1.$

My tangential approach

My 2nd approach in the My Attack Of Problem (iii) section of this query
may be re-interpreted as a tangential algorithm for identifying the square roots of $w_1.$ This means that given any messy complex expression $w$, one might identify the square roots of $w$ as follows:

  1. Identify (for example) a fourth degree equation of the form $\;[E]\;\;az^4 + bz^2 + c = 0.\;$

  2. Interpret this as a quadratic equation in $z^2,$ one of whose roots is $w.$

  3. As in my 2nd approach in the My Attack Of Problem (iii) section, $\;E,\;$ must be readily factorable into two 2nd degree polynomials.

  4. Further, each of the two polynomials must be readily solvable. This means that for each polynomial, its resultant expression $\;\sqrt{b^2 - 4ac},\;$ must be readily computable. This means that the sine and cosine of the corresponding principal Argument must be readily computable.

Note: Since there is flexibility in choosing any equation $\;E,\;$ one of whose roots is $w,$ there needs to be guidelines for designing $\;E,\;$ so that is readily factorable into two 2nd degree polynomials, each of whom is readily solvable.

My Related Questions

I am way out of my depth here, and request responses from professional mathematicians.

  1. Ignoring my tangential approach, is there a standard method of computing the square roots of such a messy complex number as $w_1.$

  2. Is my tangential approach viable? Is it a standard method? Are there guidelines for designing the corresponding helper equation $\;E$?

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  • $\begingroup$ $(1+\sqrt{1-i})^2=w_1$. I derived this result by using a CAS. $\endgroup$ Commented Dec 28, 2018 at 10:11
  • $\begingroup$ Calculators such as Mathematica. $\endgroup$ Commented Dec 28, 2018 at 10:14
  • $\begingroup$ I'm sorry I don't know the pencil and paper algorithm. The motivation of commenting is I think knowing the answer can help you a little bit. $\endgroup$ Commented Dec 28, 2018 at 10:20
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    $\begingroup$ @KemonoChen Your response was helpful. I am very ignorant of CAS. However, in this instance I am looking for a manual algorithm that may be used to attack all problems of this type. After all, if you are going to use CAS to compute $\sqrt{w_1}$, you can (perhaps) just as readily use CAS to solve problem (iii). $\endgroup$ Commented Dec 28, 2018 at 10:24
  • $\begingroup$ You might be after the elementary result that, for every $z$ complex number which is not a nonnegative real number, the square roots of $z$ are $$\pm\sqrt{|z|}\frac{z+|z|}{\left|z+|z|\right|}$$ $\endgroup$
    – Did
    Commented Dec 29, 2018 at 6:22

3 Answers 3

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Not a full solution, but an elaboration on Palka's approach. You do not have to compute $\alpha$. Following Palka, you can use the following:

$$ \;\displaystyle w_1 \;=\; \left(2 + \sqrt{\sqrt{2}}\sqrt{2 + \sqrt{2}}\right) + i\left(-1 -\sqrt{\sqrt{2}}\sqrt{2 - \sqrt{2}}\right) = \cal{{R}} + i \cal{{I}} $$ where $\cal{{R}}, \cal{{I}}$ identify the real and imaginary parts of $w_1$. Now we have the following relations: $$|w_1|^2 = \cal{{R}}^2 + \cal{{I}}^2\\ w_1 = |w_1|(\cos \alpha + i \sin \alpha) = \cal{{R}} + i \cal{{I}}\\ \sqrt w_1 = \sqrt{|w_1|} (\cos \beta + i \sin \beta) = \sqrt{|w_1|} \left(\sqrt{\frac{1 + \cos(\alpha)}{2}}+ i \sqrt{\frac{1 - \cos(\alpha)}{2}}\right) = \\ = \sqrt{\frac{|w_1| + |w_1|\cos(\alpha)}{2}}+ i \sqrt{\frac{|w_1| - |w_1|\cos(\alpha)}{2}}\\ = \sqrt{\frac{\sqrt{\cal{{R}}^2 + \cal{{I}}^2} + \cal{{R}}}{2}}+ i \sqrt{\frac{\sqrt{\cal{{R}}^2 + \cal{{I}}^2} - \cal{{R}}}{2}} $$ This means you can directly put in $\cal{{R}}, \cal{{I}}$ which are given from the original task.

Regarding Palka's last hint, writing two roots as $z_{1,2} = \frac{1}{2a}\left(-b \pm \sqrt{b^2 - 4ac}\right)$, you can now write the last line above as

$$ \sqrt w_1 = \sqrt{\frac{\sqrt{\cal{{R}}^2 + \cal{{I}}^2} + \cal{{R}}}{2}}\pm \sqrt{\frac{-\sqrt{\cal{{R}}^2 + \cal{{I}}^2} + \cal{{R}}}{2}} $$

So the two arguments under the roots are the two solutions to $\;az^2 + bz + c = 0\;$ when identifying $a= 1$, $b = -\cal{{R}}$, $c = - {\cal{{I}}^2}/4$.

So $ \sqrt w_1 = \sqrt z_1 \pm \sqrt z_2 = \sqrt z_1 \pm i \sqrt{-z_2}$ which also gives the right structure in real and imaginary parts, since both $z_1$ and $-z_2$ will be positive.

It of course remains to put in $\cal{{R}}$ and $\cal{{I}}$ and then solve the quadratic equation and I still think this will get messy and use of Wolframalpha or the like will be helpful. However, the benefit of this treatment is that it directly gives you the required structure of the solution.

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  • $\begingroup$ I gave a (longwinded) response to your answer, as a separate answer. I welcome feedback. $\endgroup$ Commented Dec 29, 2018 at 2:43
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I would like to respond to Andreas' answer. I feel that it would be too confusing to add this response as an addendum to my original question. Also, I think my response will be more legible as an answer, rather than a comment.

(1)
There seems to be a typo: in Andreas' math expression directly below
"now write the last line above as...".
I think that the first term on the right hand side should be

$\displaystyle \sqrt{\frac{\sqrt{\cal{R}^2 + \cal{I}^2} + \cal{R}}{2}}$

Andreas, if I am mistaken, please respond.

(2)
Interpreting Andreas' initial expression for $\;\sqrt{w_1}\;$ as $\;\displaystyle \pm \left( \sqrt{\frac{\sqrt{\cal{R}^2 + \cal{I}^2} + \cal{R}}{2}} \;+\; i\,\sqrt{\frac{\sqrt{\cal{R}^2 + \cal{I}^2} - \cal{R}}{2}} \right)\;,$
my reaction is: that is not necessarily accurate.

In fact, with the $w_1$ as specified in my original query, $\;\displaystyle \sqrt{w_1} \;=\; \pm \left( \sqrt{\frac{\sqrt{\cal{R}^2 + \cal{I}^2} + \cal{R}}{2}} \;-\; i\,\sqrt{\frac{\sqrt{\cal{R}^2 + \cal{I}^2} - \cal{R}}{2}} \right)\;.$

To the best of my knowledge, the only way in general to determine which of the two expressions is accurate is to let $\beta$ represent $\;(1/2) \;\text{Arg}(w_1),\;$ and then determine whether $\;\cos(\beta) < 0\;$ and also determine whether $\;\sin(\beta) < 0.$

(2) mea culpa
see Andreas' comment following this answer. Apparently, I misinterpreted his evaluation of $\cal{I}.$ Anyway, the original section (2) [above] left as is, as a reference.

(3)
As Andreas indicated, manually implementing his approach (i.e. without resorting to CAS) might not lead to a clear expression for $\;\sqrt{w_1}.$
For example, using the values of $w_1$ and $z_2,$ from my original query,
and construing that $\;w_1 = \cal{R} + i\cal{I},\;$ leads to
$\displaystyle \cal{R} \;=\; 2 + \sqrt{\sqrt{2}}\sqrt{2 + \sqrt{2}}, \;\; \cal{I} \;=\; -1 - \sqrt{\sqrt{2}}\sqrt{2 - \sqrt{2}}.$

There doesn't seem to be any way to use his method to manually derive that one of the roots of $\;\sqrt{w_1}\;$ is $z_2.$

Contrast this with (as indicated in my query) my accidental use of the tangential method, which led to the direct manual derivation of $z_2.$

(4)
When $\;a=1, \;b=-\cal{R}, \;c=-\cal{I}^2/4,\;$ the two roots of
$\;az^2 + bz + c = 0\;$ will be $\displaystyle \frac{1}{2} \left(\cal{R} \pm \sqrt{\cal{R}^2 + \cal{I}^2} \right),\;$ which doesn't seem to match his expression.

Andreas, again, if I am mistaken, please respond.

(5)
Assuming that a quadratic equation is identified, one of whose roots is $\;\sqrt{w_1},\;$ the other root to that specific quadratic equation may not be $\;-\sqrt{w_1}.\;$ Assuming that that is the case, then (to the best of my knowledge), you will have to square both of the roots to the quadratic equation, and see which square matches $w_1.$

In fact, the only quadratic equation that will have the two roots of $\;\sqrt{w_1}, \;-\sqrt{w_1}\;$ will be $z^2 - w_1 = 0.$

$\underline{\text{addendum-1}}$
After consideration, I realized that my point (2) above, although accurate, is somewhat ill considered. Construing $\;w_1$ as $\;\cal{R} + i\cal{I},\;$ setting $\;\alpha \;=\; \text{(the principle) Arg}(w_1),\;\text{\{i.e.}\; \alpha \,\in (-\pi,\pi]\}\;$ and setting $\displaystyle \beta \;=\; \frac{\alpha}{2},\;$ then
(a) $\;\sin(\alpha)\;$ will be negative iff $\cal{I}$ is negative.
(b) $\;\sin(\beta)\;$ will be negative iff $\;\sin(\alpha)\;$ is negative.
(c) $\;\cos(\beta)\;$ will always be non-negative.

$\underline{\text{addendum-2}}$
The following is an elaboration of the accidental tangential approach expressed in my original query, for manually deriving the square roots of a messy complex $w.\;$ There may well be other (? more viable ?) approaches.

Select complex $b$ and $c$ so that all of the following conditions are satisfied.

  1. Form equation $E$ as $\;z^4 + bz^2 + c = 0.$

  2. $w$ must equal $\displaystyle\;\frac{1}{2}\left(-b \pm \sqrt{b^2 - 4c}\right).$
    Although $\;\displaystyle \sqrt{b^2 - 4c}\;$ does not have to be readily manually derivable, you do have to be able to readily manually confirm that condition 2 is met.

  3. $E$ may be factored into $\;(z^2 + rz + s) \times (z^2 - rz + s)$
    where $\;s^2 = c,\;$ and $\;(-r^2 + 2s) = b.$

  4. $\sqrt{r^2 - 4s}\;$ must be readily manually derivable.

I know of no guidelines for choosing $b$ and $c$ so that all of the above conditions are satisfied.

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  • $\begingroup$ Thanks for the detailed comments. I changed the typo (1). As for (2), did you notice that I identified $-1 -\sqrt{\sqrt{2}}\sqrt{2 - \sqrt{2}} = \cal{{I}}$ ? (Note the minus-signs / compare to the original question). In (4), yes: what you wrote are exactly the two arguments of the roots. $\endgroup$
    – Andreas
    Commented Dec 30, 2018 at 15:27
  • $\begingroup$ @Andreas thanks for the feedback. I'll update (2) with a mea culpa. $\endgroup$ Commented Dec 30, 2018 at 17:16
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I have tried to give a simple approach to deal with square roots of $w_1$. Let $$a=\sqrt{2\sqrt{2}+2},b=\sqrt{2\sqrt{2}-2}$$ so that $$ab=2, a^2-b^2=4$$ and $$w_1=2+a-(1+b)i=(a+2)-i\left(1+\frac{2}{a}\right)=A-iB$$ where $A, B$ are positive real numbers.

The square roots of the above complex number $w_1$ are given by $$\pm\sqrt{\frac{\sqrt{A^2+B^2}+A} {2}}\mp i\sqrt{\frac{\sqrt{A^2+B^2} - A}{2}}$$ (signs of the real and imaginary parts are opposite to each other as imaginary part of $w_1$ is negative). Now $$A^2+B^2=(a+2)^2\cdot\frac{a^2+1}{a^2}$$ Note that $$a^2+1=3+2\sqrt{2}=(1+\sqrt{2})^2=\frac{a^4}{4}$$ and hence $$\sqrt{A^2+B^2}+A=\frac{(a+2)a}{2}+(a+2)=\frac{(a+2)^2}{2}$$ and $$\sqrt{A^2+B^2}-A=\frac{(a+2)(a-2)}{2}=\frac{b^2}{2}$$ It follows that the desired square roots are $$\pm\frac{a+2}{2} \mp i\cdot\frac{b} {2} $$ The particular number $w_1$ was choosen to have specific relation between its real and imaginary parts which led to some simplification.

More generally the square roots of a complex number can be computed algebraically using the notion of square root of a real number and we can expect to a get a radical expression which may or may not be simplified further.

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  • $\begingroup$ I appreciate your interest and effort. I have just bookmarked this question, and in the next couple of days I will be studying your answer. $\endgroup$ Commented Aug 25, 2021 at 21:16
  • $\begingroup$ +1: What can I say. Very elegant, and just what I was looking for. $\endgroup$ Commented Aug 26, 2021 at 9:58

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