Find all real numbers x satisfying Re[(1+ix)^3]=Im[(1+ix)^3]
Find all real numbers $x$ satisfying $\ \text{Re}\big[(1+ix)^3\big]=\text{Im}\big[(1+ix)^3\big]$.
Explanation of Method
Expand $(1+ix)^3$ by the binomial theorem, separate real and imaginary parts (each a real polynomial in $x$), set them equal, and solve the resulting real cubic.
Step 1 · Expand $(1+ix)^3$
$$ (1+ix)^3=1+3ix+3(ix)^2+(ix)^3=1+3ix-3x^2-ix^3 $$
$$ =(1-3x^2)+i(3x-x^3) $$
so $\text{Re}=1-3x^2$ and $\text{Im}=3x-x^3$.
Step 2 · Set the real and imaginary parts equal
$$ 1-3x^2=3x-x^3 \quad\Longrightarrow\quad x^3-3x^2-3x+1=0 $$
Step 3 · Find a rational root
Testing $x=-1$: $\ (-1)^3-3(-1)^2-3(-1)+1=-1-3+3+1=0$ ✓. So $(x+1)$ is a factor.
Step 4 · Factor and solve the remaining quadratic
$$ x^3-3x^2-3x+1=(x+1)(x^2-4x+1) $$
The quadratic $x^2-4x+1=0$ gives $x=\dfrac{4\pm\sqrt{16-4}}2=2\pm\sqrt3$.
Step 5 · Conclusion
$$ \boxed{x=-1, \quad x=2+\sqrt3, \quad x=2-\sqrt3} $$
Verification
| $x$ | $1-3x^2$ | $3x-x^3$ |
|---|---|---|
| $-1$ | $-2$ | $-2$ |
| $2+\sqrt3\approx3.732$ | $\approx-40.78$ | $\approx-40.78$ |
| $2-\sqrt3\approx0.268$ | $\approx0.785$ | $\approx0.785$ |
All three roots were confirmed symbolically by factoring $x^3-3x^2-3x+1=(x+1)(x^2-4x+1)$ exactly.
Result
Separating a binomial power of $1+ix$ into real and imaginary parts always yields two real polynomials in $x$; equating them is just solving a real polynomial equation.