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:
-
Identify (for example) a fourth degree equation of the form
$\;[E]\;\;az^4 + bz^2 + c = 0.\;$ -
Interpret this as a quadratic equation in $z^2,$ one of whose roots is $w.$
-
As in my 2nd approach in the My Attack Of Problem (iii) section,
$\;E,\;$ must be readily factorable into two
2nd degree polynomials. -
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.
-
Ignoring my tangential approach, is there a standard method
of computing the square roots of such a messy complex number as $w_1.$ -
Is my tangential approach viable? Is it a standard method? Are there
guidelines for designing the corresponding helper equation $\;E$?
Best Answer
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.