Which real number pairs (x, y) fulfill equation $(x+iy)^3=1$

Question

This is an exercise question from my first college math class, and I am having hard time even understanding what it means or wants me to solve. It's probably quite simple but I never studied that much mathematics in high school. Any help would be appreciated. Original question is not in English, but I am trying my best to translate it:

"Determine by solving a pair of equations all such real number pairs (x, y) which fulfill the equation below."

$(x+iy)^3=1$

Answer 1

There are several ways of going about with this. The most straightforward is to just open the parentheses: $$ \left(x+iy \right)^3 = x^3 + i3x^2 y - 3xy^2 - i y^3 $$ Let's group the terms so that we distinguish between the terms that are multiplied by $i$ from those that are not: $$\tag{1} \left(x+iy \right)^3 = \left[ x^3 - 3xy^2 \right] + i\left[ 3x^2 y - y^3 \right] $$ We know that the expression should be equal to $1$, but this also means that it should be equal to $1+i\cdot0$. Therefore, the first bracket on the right-hand side of Equation (1) should be equal to 1 and the second bracket should be equal to zero: $$ \left\{ \begin{array}{ccccc} x^3& - &3xy^2 &=& 1 \\ 3x^2 y& - &y^3 &=& 0 \end{array} \right. $$ This pair of equations is not very difficult to solve. Can you continue from here?

Answer 2

The easiest way to compute $x$ and $y$ is through polar coordinates: you can write $x+iy$ as $re^{i\theta}$ where $r>0$ and $\theta$ is real. This gives $r^{3}e^{3i\theta}=1$. Taking modulus we get $r=1$. We then get $3 \theta=2n\pi$ where $n$ is an interger. Going back to $x$ and $y$ we get $x=\cos (\frac {2n \pi} 3) $ and $y=\sin (\frac {2n \pi} 3)$ where $n$ is an interger.

Answer 3

Observe that $$\Im(x+iy)^3=3x^2y-y^3=(3x^2-y^2)y=0.$$

Then

• $y=0\to x^3=1$ and $x=1$,

• $y=\pm\sqrt3x\to x^3(1\pm\sqrt3i)^3=-8x^3=1$ and $x=-\frac12$.

Answer 4

$(x+iy)^3-1^3=0;$

$\small{(x+iy-1)((x+iy)^2+(x+iy)1+1)=0}$

1)$x-1+iy=0$; $x=1$; $y=0$;

2)$x^2+2ixy-y^2+x+iy+1=0$;

a)$x^2-y^2+x+1=0$;

b)$2xy+y=0$;

Note: $y \not =0$ (why?)

b') $x=-1/2$;

a') $1/4-y^2+1/2=0$;

$y^2=3/4$; $y=\pm (1/2)√3$.

Hence $(-1/2,\pm(1/2) √3)$.

Answer 5

For all $x+yi$ actually can be expressed as a complex number $z$. Then, we are finding the roots of $z^3=1$. So the roots are $e^0,e^{2\pi i/3},e^{4\pi i/3}$, which is $1+0i,-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i,-\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}i$