3
$\begingroup$

Solving a second degree complex polynomial: -$x^2 + \sqrt3 x$ + i.

Now taking the standard form of equating the coefficient of a=1:$\;\;$ $x^2 -\sqrt3 x$ - i.

Solving using the quadratic formula, the discriminant is : 3 + 4i. Need to find the root (square root) for the same. Assuming variable y for denoting the roots of the quadratic equation:

$y^2$ = 3 + 4i $\;\;\;$$\;\;\;\;$-- equation (a).

I want to pursue both algebraic and trigonometric solutions.

Trigonometric solution:

Taking the polar form, the representation of 3+4i is : $5(\cos(arctan(4/3)) + \sin(arctan(4/3))$.

Is there any simplification possible, so that the table for arctan need not be consulted?

The expression of interest is: $\cos(\theta) = 3/5, \sin(\theta)= 4/5$. I tried using various trigonometrical identities, but failed.

Based on response by MyGlasses, the approach is :

5($\dfrac{1}{\sqrt(1+x^2)} + x.\dfrac{1}{\sqrt(1+x^2)})$, where $x = \arctan(4/3)$, $\;\;\;$-\pi.

=> 5($\dfrac{1+x}{\sqrt(1+x^2)})$

May be I have taken it totally wrong, but seems that with the current form the application of De-Moivre's theorem is not possible. Also, there re two more errors: (i) x is taken as an angle, and the trig functions are not rational functions of the angle itself, (ii) x is only a constant value, that would not help in finding all the roots, as this would lead to : $3+4i = 5.$(some constant fraction).

I am simply taking the angle, let $\theta$ = arctan(4/3) & taking the cosine and sine for the same :

5($\cos(\arctan$(4/5) + $\sin(\arctan$(3/5) ) $\;\;\;\;$-- equation (b)

As need for polar form, the explicit representation of angles also (apart from modulus), rather than just sine and cosine expressions. Otherwise, can point to multiple solutions, rather than a single one. So, find out the values of the two angles:

Now, directly use the trigonometric identities for $\cos(\theta)$ = 2$\cos^2(\dfrac{\theta}{2})$ -1 , given $\cos(\theta)= \frac{3}{5}$.

=> $\cos(\dfrac{\theta}{2})$ = $\sqrt(\dfrac{1+\cos(\theta)}{2})$ = $\sqrt(\dfrac{4}{5})$, with only positive sign being considered for the square root, as the value of cosine lies in the first quadrant.

Similarly, $\cos(\theta)$ = 1-2$\sin^2(\dfrac{\theta}{2})$ => $\sin(\dfrac{\theta}{2})$ = $\sqrt(\dfrac{1-\cos(\theta)}{2})$ = $\sqrt(\dfrac{1-\frac{3}{5}}{2})$ = $\sqrt(\dfrac{1}{5})$, with only positive sign being considered for the square root, as the value of sine lies in the first quadrant.

Now, the square root of the polar form expression is given as follows, if we consider the first quadrant solution only :

$\sqrt(5)(\cos(\dfrac{\theta}{2}) + i\sin(\dfrac{\theta}{2}) ) \implies sqrt(5)(\sqrt(\dfrac{4}{5}) + i\sqrt(\dfrac{1}{5}) )$ $=> 2 +i$

Otherwise, generally speaking, can add $2\pi \frac{k}{m}, k,m \in \mathbb{Z}, (k $ for the actual root (i.e. first, second for square root, corresponding to $k=0,1,$ respectively, and $m$ for the number of roots) to the angle in eqn .(b); and the principal a particular root has been already been found out (half angle, as square root means (original angle)/2, and k=1, for $2k\pi$); so can simply add $\pi (= 2.\pi\frac{1}{2})$ to the principal particular root to find another root.
Additionally, can see details at : here in MSE, and external link.
Also, there is definition of 'principal root' given at (although, a more valid definition would be first finding any particular root, and then taking advantage of symmetry to find the others) wolframalpha :

For complex numbers z, the root of interest (generally taken as the root having smallest positive complex argument) is known as the principal root.

However, at this link the site of wolframalpha contradicts by taking $-15^0$ as the smallest positive root (particular root) for the given equation of finding cube root of $(1-i)$. So, would continue using the term: "a particular solution" instead.

$\implies$ So, another (second) root will lie in the 3rd quadrant, symmetrically dividing the $360^{\circ}$ argand plane. Note that this will be diametrically opposite to the particular principle root (2+i), hence (-2-i).


As an aside, consider the different notations associated with two complex roots of a quadratic equation, that directly show the lack of relevance of the 'principal root'.
In the book titled, Complex Analysis and Applications, 2nd edn., by Allan Jeffrey; it is stated on page #20,21 about the different possible values of the discriminant D for a quadratic equation.

I imagine in the below paras., the author wants to reinforce the fact that the square root of a complex value is a multi-valued "function", as exemplified by link 1 & link 2:>>>

The book states 3 different cases, with the first two concerning the usual ones (D being a positive real, and negative real respectively).

The 3rd case refers to the case when D is an imaginary quantity. The book states that for this case, the negative sign before the discriminant (D) is removed, as the imaginary quantity will generate both positive and negative values.


Algebraic solution:

Let, $y = a +bi$, with $a,b$ being real valued. $\;\;\;\;\; (a+bi)^2 = 3+4i \implies a^2 -b^2 +2abi = 3 +4i$

=> $a^2 - b^2 = 3 \;\;$ -(i) $\;\;\;\;\;\;\;\; ab = 2 \;\;$ -(ii)

Substituting from (ii) to (i) for any variable (let, $a$) does lead to a quartic equation:

$b^4 + 3b^2 -4 =0 $, now have a new variable $c$, s.t. $c$ = $b^2$

$c^2 +3c -4 =0,$ --equation (iii)
The roots are -4, 1.

So, $(b^2 = -4)$ $\cup (b^2 = 1) $ => $(b = \pm\sqrt(-4)) \cup (b = \pm 1)$

=> $(b = \pm 2i) \cup (b = \pm 1)$

$b = \{2i, -2i, 1, -1\}$

On substituting the 4 values of $b$ in eqn. (iii) from the available possible value set; get that only $b = \pm$1 work for $b$ being real.

=> a = $\pm$2.

Hence, the complex root (y) values are : (2 +i), (-2-i).

So, the original eqn. has roots as:

x = $\dfrac{\sqrt3 \pm y' }{2}$, where y' = 2 + i.

$\endgroup$
  • $\begingroup$ Sorry. I did misunderstand the question. $\endgroup$ – Nosrati Sep 25 '17 at 8:41
  • $\begingroup$ The title is doubly misleading. You might want to replace it by "Roots of complex quadratic polynomial". $\endgroup$ – Did Sep 25 '17 at 9:21
  • $\begingroup$ In the end, what are you asking? Whether the solutions of cos(θ)=3/5, sin(θ)=4/5, are known? Note that, to solve (a+bi)^2=3+4i, one can first try small integer values of a and b... or use the half-angle formula for the tangent. $\endgroup$ – Did Sep 25 '17 at 9:23
  • $\begingroup$ @Did: I cannot understand the expression : "first try small integer values of a and b". It is seemingly numerical approximation approach. Also, how it would apply here. Plus, if such simple calculation needs numerical approximation; then really new to me. $\endgroup$ – jiten Sep 25 '17 at 12:29
  • 1
    $\begingroup$ You are confusing $x$ and $y$ big time... So... the solutions of $y^2=3+4i$ are $y=2+i$ and $y=-2-i$ hence the roots of $x^2−\sqrt3 x - i.$ are $x=\frac12(\sqrt3+y)=\ldots$ $\endgroup$ – Did Sep 25 '17 at 18:17
1
$\begingroup$

There’s a number-theoretic way to approach the question of determining $\sqrt{3+4i\,}$ as well.

In field theory, the Norm is a multiplicative map from a field $K\supset\Bbb Q$ that’s finite-dimensional over $\Bbb Q$, down to $\Bbb Q$. If you want a factorization of $z\in K$, it may help to look at the $\Bbb Z$-factorization of $\text{Norm}(z)$. For $K=\Bbb Q(i)$, the Norm of $a+bi$ is $a^2+b^2$. Here, the Norm of $3+4i$ is $25$, a square integer, suggesting that $3+4i$ may be a square as well. No guarantees here, but you do have the two obvious “primes” of Norm $5$, namely $1+2i$ and $2+i$, and it makes sense to square them to see whether the result may be of help with our question. The first of these squares to $-3+4i$, probably of no help, but the square of the second is $3+4i$.

Note that the idea of using the Norm was not a technique for calculating $\sqrt{3+4i\,}$, but rather a search strategy, not guaranteed to work. If instead you had been interested in $\sqrt{4+3i\,}$, there would have been additional steps, requiring you to venture outside the ring $\Bbb Z[i]$.

$\endgroup$
  • $\begingroup$ Although, I am novice to Algebraic number theory (ANT), but feel you refer to Gaussian ones. Also, in your last para., you must have meant that Norm loses info. about the Gaussian integers. But, the difference from $\sqrt{3+4i}$ to $\sqrt{4+3i}$ is not clear at all, in terms of going outside Algebraic numbers even. Would request that at least some reference be given. $\endgroup$ – jiten Jan 1 '18 at 2:41
  • 1
    $\begingroup$ Although it’s true that Norm loses information, I was not saying that at all. The difference between $3+4i$ and $4+3i$ is simply that the former is $(2+i)^2$, while the latter is $i(2-i)^2$. You discover this by trial division by $2+i$ and $2-i$. So to get $(4+3i)^{1/2}$, you need to adjoin a square root of $i$, for instance $(1+i)/\sqrt2$, not a Gaussian number. No reference, it’s just hand computation. $\endgroup$ – Lubin Jan 1 '18 at 17:27
  • $\begingroup$ I request an advice, that need as a novice seeking idea about ANT. Is the hand computation having a clear-cut way to ascertain when to stop? I have seen examples where GAP is used to get roots. How is hand computation better than GAP. I mean 'better' in terms of developing better intuition, which is possible only if know when to stop - which normally (actually, better term is : 'known to me') is - being prime in Gaussian/algebraic integers. So, if not prime, then should go on trying? Sorry, if asked without poking enough, but seemingly have to poke a lot. $\endgroup$ – jiten Jan 1 '18 at 23:10
  • 1
    $\begingroup$ Sorry, but I don’t know anything about GAP. For the most part, I have found machine computation less good than the direct kind with pencil and paper. $\endgroup$ – Lubin Jan 2 '18 at 2:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.