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.
Best Answer
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 $\Bbb Z[i]$.