Roots of a complex equation

Question

For , $ f(z) = z^3 + iz^2 + iz -1$ . Prove $f(z) =0$ has all of its roots as imaginary and the sum of two of the roots is zero.

This is how is proceeded ,

Given polynomial can be written as ,

$z^3 + iz^2 + iz + i^2 = 0$

$ (z^2 +i)(z+i) =0$

Hence we have ,

$z=-i$ or $z^2 + 0z + i= 0$

Clearly from the quadratic we can check that sum of two of its roots is zero. How can we prove that all of its roots are imaginary ?

Answer 1

Note that one root $z_1=-i$ is already imaginary. Now, note that $$z^2=-i $$ $$\implies z^2=e^{i\frac{3\pi}2}$$ $$\implies z= \pm e^{i\frac{3\pi}4} = \mp \frac1{\sqrt 2} \pm \frac{i}{\sqrt 2}$$ both imaginary roots.

Answer 2

Now, write $$z^2+i=z^2-\left(\frac{1-i}{\sqrt2}\right)^2,$$ which gives $$z_1=\frac{1}{\sqrt2}-\frac{1}{\sqrt2}i$$ and $$z_2=-\frac{1}{\sqrt2}+\frac{1}{\sqrt2}i.$$

Answer 3

Another basic method to do this:

Write $z=x+iy$, $x,y\in\mathbb R$ so that

$z^2=-i\implies (x+iy)^2=-i\implies x^2-y^2+2ixy=-i$. Now equating real and imaginary parts,

$x^2-y^2=0$ and $2xy=-1$

$\implies x=\pm y$ and $2xy=-1$.

Plug $x=-y$ in the other one to get $x^2=1/2$ i.e. $x=\pm 1/\sqrt{2}$ and $y=\mp 1/\sqrt{2}$. Plugging $x=y$ in the other one will lead to imaginary values of $x$ and $y$, hence this is rejected.

So $z=\pm 1/\sqrt{2 }\mp i/\sqrt{2}$ are the roots along with $z=-i$.