The roots of the cubic equation, $2x^3+px^2-(3+5i)x+q=0$ are $x\in \{ i , \frac{1}{k}, -1-k\} $ Find $p,q$ and $k$. 

The roots of the cubic equation, $2x^3+px^2-(3+5i)x+q=0$ are $x\in \{ i , \frac{1}{k}, -1-k\} $ 
Use this information to find $p,q$ and $k$, where $k$ is real.


What I have done:
Consider $p(x)=2x^3+px^2+(-3-5i)x+q=0$
Since $x=i$ , $x=\frac{1}{k}$ and $x=-1-k$ are roots we will try subbing them in and seperating imaginary and real parts and solving simultaneously hopefully!
Starting of with $x=i$
$$\Longrightarrow p(i)=2(i)^3+p(i)^2+(-3-5i)i+q=0$$
$$ \Leftrightarrow -2i -p +5 - 3i +q=0$$
$$ \Leftrightarrow q-p+5-5i=0$$
$$ \Leftrightarrow (q-p+5)+i(-5)=0$$
So one equation we have is $q-p+5=0$ 
Next $x=\frac{1}{k}$
$$\Longrightarrow p(\frac{1}{k})=2(\frac{1}{k})^3+p(\frac{1}{k})^2+(-3-5i)\frac{1}{k}+q=0$$
$$\Leftrightarrow \frac{2}{k^3} + \frac{p}{k^2} - \frac{3}{k} - \frac{5i}{k}+q=0$$
$$ \Leftrightarrow (\frac{2}{k^3} + \frac{p}{k^2} - \frac{3}{k} +q) + i(\frac{-5}{k})=0$$
So another equation we have is $\frac{2}{k^3} + \frac{p}{k^2} - \frac{3}{k} +q=0$ since $\frac{-5}{k} \neq 0$
The last root we have is $x=-1-k$ 
$$\Longrightarrow p(-1-k)=2(-1-k)^3+p(-1-k)^2+(-3-5i)(-1-k)+q=0$$
$$ \Leftrightarrow -2(k+1)^3+p(k+1)^2+3+3k+5i+5ki+q=0 $$
$$ \Leftrightarrow (-2(k+1)^3+p(k+1)^2+3+3k+q)+i(5+5k) =0 $$
So 2 other equations we have are $-2(k+1)^3+p(k+1)^2+3+3k+q=0$ and $5+5k=0$
Hence the 4 equations we have are 
$$q-p+5=0 $$
$$\frac{2}{k^3} + \frac{p}{k^2} - \frac{3}{k} +q=0$$
$$-2(k+1)^3+p(k+1)^2+3+3k+q=0$$
$$5+5k=0$$
For the last equation $k=-1$ and subbing this into the third equation I get $q=0$
and subbing that into the second equation I get $p=1$ but subbing $q$ and $p$ into $q-p+5$ does not make it true so where did I go wrong? I also do not think I have made any algebraic error. 
 A: Via long division of the polynomial by $x-i$ we get
$$2x^3-px^2-(3+5i)x+q=(x-i)(2x^2+(p-2i)x+(-1+(p-5)i)$$
Immediately we run into an error in your working, since $q=i+(p-5)$, not $q-p+5=0$. Rewinding to the line where this false relation is introduced:
$$(q-p+5)+i(-5)=0$$
we see that going from this to $q-p+5=0$ implies that $-5i=0$, which is absurd.
The other relations you derive (with $x=\frac1k,-1-k$) are also flawed. The correct equations those incorrect relations stem from:
$$\left(\frac{2}{k^3} + \frac{p}{k^2} - \frac{3}{k} +q\right) + i\left(\frac{-5}{k}\right)=0$$
$$(-2(k+1)^3+p(k+1)^2+3+3k+q)+i(5+5k) =0$$
are of the form $a+bi=0$, but you cannot say that $a=0$ or $b=0$ in this case as you did because $a$ and $b$ can be complex here. They need to be restricted to the real numbers in order for you to conclude that $a=b=0$.
We shall solve for $p,q,k$ anyway. The remaining roots after the long division are all real, so the coefficients of the quotient polynomial must be too. $p-2i$ real implies $p=a+2i$ with $a$ real. $-1+(p-5)i=-3+(a-5)i$ is also real, so $a=5$, $p=5+2i$ and our remaining quadratic equation is
$$2x^2+5x-3=0$$
which turns out to have roots $\frac12$ and $-3$, so $k=2$. $q$ works out to be $i+(5+2i-5)=3i$.
