Algebraic solution of complex equation

Question

For solving algebraically any complex equation involves two components for the real & imaginary parts. Let the real part be - $a$, imaginary part - $b$. For the complex equation $$x^3 = 1-i $$

Substituting $x = a +bi$, we get: $$(a+bi)^3 = 1 - i.$$ Expanding the l.h.s. : $$a^3 -(b^3)i + 3(a^2)bi -3(b^2)a$$ Equating real and imaginary parts, we get: $$ a^3 -3(b^2)a = 1 \tag{1}$$
$$ b^3 -3(a^2)b = 1 \tag{2}$$

Factoring $(1)$, $(2)$ to get some roots in the process: $$a(a^2 - 3(b^2)) = 1 \tag{1'}$$ $$b(b^2 - 3(a^2)) = 1 \tag{2'}$$

From $(1')$, the roots are possibly given by equations below:

$$a = 1 \tag{3}$$ $$ - \text{or} - $$ // 'or' is used in logical or sense, i.e. either or both

$$ a^2 - 3(b^2) = 1 \tag{4}$$ Substituting $a = 1$ in $(4)$, we get: $b = 0$; which cannot be a possible solution, as the right side of question has $b = -1$. Hence, $a \ne 1$ also.

Next, trying for the possible value from $(2')$. $$ b = 1 \tag{5} $$ $$ -\text{or}- $$ // 'or' is used in logical or sense, i.e. either or both

$$ b^2 - 3(a^2) = 1 \tag{6} $$ Substituting $b = 1$ in $(6)$, we get: $$ a = 0; $$ which cannot be a possible solution, as the right side of question has $a = 1$. Hence, $b \ne 1$ also.

I am unable to solve it further, as no solution emerges from the two equations - $(1')$, $(2')$.

Answer 1

No, this is wrong. $a(a^2 - 3 b^2) = 1$ does not imply $a=1$ or $a^2-3b^2=1$. For example, you might have $a = 2$ and $a^2 - 3 b^2 = 1/2$.

It's usually not helpful to look at real and imaginary parts in an equation such as this. A better idea would be to use the polar representation: $x = r e^{i\theta}$, $1-i = \sqrt{2} e^{-i\pi/4}$.

Answer 2

Hint: $$a^3-3b^2a=b^3-3a^2b=1 \implies a^3-b^3+3a^2b-3b^2a=0$$ which implies $$(a-b)(a^2+ab+b^2)+3ab(a-b)=0 \implies (a-b)(a^2+4ab+b^2)=0.$$ As such, $a=b$ or $a^2 + 4ab+b^2 = a^2+4ab+4b^2-3b^2 = 0 \implies a = (-2 \pm \sqrt{3})b.$

If $a=b$ then $a^3-3b^2a = -2a^3=1 \implies \color{red}{a=b=-2^{-1/3}}$.

I leave the calculations for $a=(-2 \pm \sqrt{3})b$ to you.

An alternative way of computing the remaining roots (as also suggested by some other people) is given below.

Let $x_0 = -2^{-1/3}-2^{-1/3}i=-2^{-1/3}(1+i)$. We just found out that $x_0^3 = 1-i$. Now $$x^3 = 1-i \implies \left(\frac{x}{x_0}\right)^3 = \frac{1-i}{x_0^3}=1.$$ Let $y = x/x_0$. We are looking for the solution of $y^3-1=0$. But $$y^3-1 = (y-1)(y^2+y+1)=0 \implies y=1 \quad \text{ or }\quad y^2+y+1=0. $$ Now $y=1 \implies x=x_0$, and $$y^2+y+1 = 0 \implies y = \frac{x}{x_0}= \frac{-1\pm \sqrt{3}i}{2}.$$ As such, the remaining cube roots of $1-i$ are $$-2^{-1/3}(1+i)\left(\frac{-1\pm \sqrt{3}i}{2}\right).$$

Answer 3

Hint: You can use polar form to find one root, then back into an algebraic solution.

$1-i= 2^{1/2}e^{-i\pi/4}$, so $(2^{1/6}e^{-i\pi/12})^3=1-i$ (there are other solutions as well).

Now write one solution $\sqrt[6]{2}\cos \pi/12 - i\sqrt[6]{2}\sin\pi/12$ in algebraic terms, then divide the appropriate linear factor from the original equation $z^3 - 1 + i =0$. That's not going to be easy.

Then you have a quadratic factor remaining, which gives its roots via the quadratic formula.

It's not too hard to show $\sin\pi/12=(\sqrt{6}-\sqrt{2})/4$ and the cosine formula is similar but with "$+$" instead of "$-$". These come from considering $\sin(\pi/3-\pi/4)$.

Answer 4

Since $1-i= \sqrt 2 (\cos315^\circ + i\sin315^\circ),$ the cube roots of $1-i$ must be $$ 2^{1/6} (\cos(105^\circ+n120^\circ) + i\sin(105^\circ+n120^\circ)) $$ where the only values of $n$ we need to consider are $0,$ $1,$ and $2.$

If we can believe the tables on this page, we have $$ \cos105^\circ = -\frac 1 4 (\sqrt 6 - \sqrt 2\,) \quad \text{and} \quad \sin(105^\circ) = \frac 1 4 (\sqrt6 + \sqrt 2\,). $$ (The page gives $\cos15^\circ = \dfrac 1 4 ( \sqrt 6 + \sqrt 2\,)$ and $\sin15^\circ = \dfrac 1 4 (\sqrt 6 - \sqrt 2\,).$ If you consider $\dfrac 1 2 = \sin30^\circ = 2\sin15^\circ\cos15^\circ,$ then this appears to make sense. And then $\cos105^\circ = -\sin15^\circ$ and $\sin105^\circ = \cos15^\circ$ since $105 = 15 +90.$)

So we should have $$ \sqrt 2\,\left( -\frac 1 4(\sqrt6-\sqrt2\,) + i \frac 1 4( \sqrt6+\sqrt 2\,) \right)^3 = 1-i. $$

Answer 5

Here is another solution, which you won't like because it is fundamentally geometric. It may not look geometric to you, but I only found it by looking at the picture.

Note that $(-1-i)^3=2-2i$. Therefore a cube root of $1-i$ is $(-1-i)/\sqrt[3]2$. The others are gotten by multiplying this answer by $\omega$ and $\omega^2$, where $\omega=\frac12(-1+\sqrt{-3})$.

(Thanks fly out to Michael Hardy for the correction.)

Answer 6

Someone has pointed out that one solution of $x^3 - (1-i) =0$ is $\dfrac{-1-i}{\sqrt[3]2}.$

Therefore $x+\dfrac{1+i}{\sqrt[3]2}$ is a factor of $x^3 - (1-i).$ So we can do long division:

$$ \begin{array}{cccccccccc} & & & & x^2 & - & \dfrac{1+i}{\sqrt[3]2}x & + & i\sqrt[3]2 \\ \\ x + \dfrac{1+i}{\sqrt[3]2} & \Big) & x^3 & + & 0x^2 & + & 0x & - & (1-i) \\ & & x^3 & + & \dfrac{1+i}{\sqrt[3]2} x^2 \\ \\ \\ & & & & -\dfrac{1+i}{\sqrt[3]2} x^2 & & & - & (1-i) \\ & & & & -\dfrac{1+i}{\sqrt[3]2} x^2 & - & i x \sqrt[3]2 \\ \\ \\ & & & & & & ix\sqrt[3]2 & - & (1-i) \\ & & & & & & ix\sqrt[3]2 & - & (1-i) \\ \\ \\ & & & & & & & & 0 \\ \hline \end{array} $$ Thus we have found that $$ x^3-(1-i) = \left( x + \frac{1+i}{\sqrt[3]2} \right) \left( x^2 - \frac{1+i}{\sqrt[3]2} x + i\sqrt[3]2 \right). $$ The other two solutions are now solutions of a quadratic equation. If we write the equation as $ax^2+bx+c=0$ then we have $a=1,$ $b= - \dfrac{1+i}{\sqrt[3]2},$ and $c=i\sqrt[3]2,$ and $$ b^2-4ac = -3i\sqrt[3]2. $$ So what is $\pm\sqrt{b^2-4ac~} \text{ ?}$ We just saw that $(1+i)^2 = 2i,$ i.e. $\pm\sqrt{2i} = \pm(1+i),$ so that gives us $\pm\sqrt{b^2-4ac~}.$

Answer 7

Square both sides:

$(x^2)^3=-2i=2(-i)=2i^3$

Then

$x^2=(2^{1/3})i$

for one root.

Now just divide the $x^3$ from your original equation by $x^2$ as given above and you have a root for $x$! You then get the others by multiplying by the complex cube roots of unity.