Interval in complex

Let roots are $p,q+im,r-im$ .The imaginary parts should cancels out because sum of the roots is zero.

$p+q+im+r-im=0 \\ \implies p+q+r=0 \\ \implies p=-q-r$

new form of the roots are $-q-r,q+im,r-im$ .

now form the equation above

$-(q+r)(q+im) - (q+r)(r-im)+(q+im)(r-im)=3+2i \\ \implies -(q+r)^2 +qr +m^2=3 \,\text{and} \, m(r-q)=2 \,\,\text{..........1st part}$

$-(q+r)(q+im)(r-im)=-(-1+ai)=1-ai \\ \implies -(q+r)(qr+m^2+im(r-q))=1-ai$

So, $-(q+r)(qr+m^2)=1 \, \text{and}\, -(q+r)m(r-q)=-a$

As from above $m(r-q)=2$ we get $q+r=\dfrac{a}{2}$

Then $qr+m^2=-\dfrac{2}{a}$

Now putting this values in the first part:

$-\dfrac{a^2}{4}-\dfrac{2}{a}=3 \\ \implies a^3+12a+8$

Let $f(a)= a^3+12a+8 \\ \implies f'(a)=3(a^2+4)>0$

So, the above function is always increasing.

Now $f(0)=8$ and $f(-1)=-5$ .So,there is exactly one root exists in $(-\infty,\infty)$ .But the smallest range in the integer form is $(-1,0)$ .

So ,solving the cubic equation we can get the only value of $a=-0.644$