For solving the complex equation $x^3$ = 1-i, approached through two ways:
(i)Trigonometric solution:
$x^3$ = $\sqrt(2)(\cos(\dfrac{-\pi}{4}) + i\sin(\dfrac{-\pi}{4}))$
In the generalized form, with $2k\pi$ as the interval for repetition of the roots.
$x^3$ = $\sqrt(2)(\cos(8k-1).\dfrac{\pi}{4}) + i\sin(8k-1).\dfrac{\pi}{4}))$
By De-Moivre's theorem:
$x = (2^{1/6})(\cos(8k-1)\dfrac{\pi}{12}) + i\sin(8k-1)\pi/12))$
Taking k=0, 1, 2; the 3 unique roots are:
k=0:
$x_1 = (2^{1/6})(\cos(-\pi/12)+i\sin(-\pi/12)) => (2^{1/6})(\cos 15^{\circ} - i\sin 15^{\circ})$
k=1:
$x_2 = (2^{1/6})(\cos(7\pi/12) + i\sin(7\pi/12))$ => $(2^{1/6})(\cos(105^{\circ}) + i\sin 105^{\circ})$=> $(2^{1/6})(-\sin15^{\circ} + i\cos15^{\circ})$
k=2:
$x_3 = (2^{1/6})(\cos(15\pi/12) + i\sin(15\pi/12))$ $=> (2^{1/6})(\cos(225^{\circ}) + i\sin(225^{\circ}))$ $=> (2^{1/6})(\dfrac{-1-i}{\sqrt(2)})$
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(ii) Algebraic solution: Have first substituted x = a+bi, found the corresponding real and imaginary parts' equations as :
$a^3$ - 3a$b^2$ =1 --(x)
$b^3$ - 3b$a^2$ =1 --(y)
Equating (x) and (y), got another equation:
$a^3$ - 3a$b^2$ = $b^3 - 3ba^2 $ --(z)
=> $a^3$ - 3a$b^2$ - $b^3 + 3ba^2$ = 0
Avoided the solving of cubic, and took the approach of breaking up the cubic equation into a product of a linear and a quadratic equation.
The form achieved is : (a-b)($a^2$+4.a.b+$b^2$) = 0 --(d)
The 3 roots are obtained as:
(i) a=b, substituting this into (x) or (y), get the value as $a_1=b_1= (-\dfrac{1}{2})^(\dfrac{1}{3})$ = -0.79370
(ii) solving the quadratic equation, get the roots as:
$a_2$ = (-2 + $\sqrt(3)$).b --(1)
$a_3$ = (-2 - $\sqrt(3)$).b --(2)
Substituting (1) into (x), get the value for $b_2=\sqrt[3]{\dfrac{(5 +3\sqrt3)}{8}}$ = 1.084;
which yields value of $a_2$ =(-2 + $\sqrt3)(\sqrt[3]{\dfrac{(5 -3\sqrt3)}{8}})$ --(3)
=>(-0.2679491924)(1.084)
=> -0.29051
The root pair is : (-0.29051, 1.084)
Substituting (2) into (x),
get the value for $b_3=\sqrt[3]{\dfrac{(5-3\sqrt(3))}{8}}$ = $\sqrt[3]{-0.02451}$ = -0.29051;
which yields value of $a_3$ =(-2 -$\sqrt3$)(-0.29051) --(4)
=> 1.084
This root is principal root: (1.084, -0.29051)
It can be checked by comparing the values of $a_1, b_1, a_2, b_2, a_3, b_3$ with trigonometrical solution; that match is there. I mean that the imaginary and real parts of the roots as given by 3 trigonometrical solutions and 3 algebraic solutions concur. On checking with the wolframalpha site, it is seen it is the case that for all 3 roots.
I want to ask an unrelated but an important question that concerns the equality of $a_2$ to $b_3$, & $b_2$ to $a_3$. Is it obvious from the quadratic equation in (d), about the equality of the parts of roots. Means is there any way to tell in advance by looking at the quadratic equation about such equality.
As an addendum to this problem- Is it possible to find the value of $cos (n^{\circ})$ or $sin (n^{\circ})$, where n is value of any angle, say $15^{\circ}$ by algebraic solution. It is somewhat important to get the solution as equating the principal root and the corresponding algebraic solution (one lying in 4th quadrant) don't help; as the answer is: $\cos15^{\circ}$ = $\frac{\sqrt6 + \sqrt2}{4}$.
