Find all the complex solutions to the equation $$iz^2+(3-i)z-(1+2i)=0$$
I've tried to solve this equation with two different approaches but in both cases I couldn't arrive to anything.
1) If $P(z)=iz^2+(3-i)z-(1+2i)$, then the original problem is equivalent to finding the roots of $P$. If I consider the polynomial $P(z)\overline{P}(z)$, then this polynomial has integer coefficients, which I thought it could make things a little bit easier.
So $P(z)\overline{P}(z)=x^4-3x^3+6x^2-10x+5$, I didn't know what to do after this.
2) I've tried to solve the equation directly, $$iz^2+(3-i)z-(1+2i)=0$$ if and only if $$z(iz+3-i)=1+2i$$
The number of the left equals the number of the right if and only if they have the same module and argument. At this point I got stuck trying to actually calculate the argument and modulus of the product on the left side of the equation.
I would appreciate any help, thanks in advance.
Best Answer
Hint: It is a quadratic equation with $a=i$, $b=3-i$, and $c=-(1+2i)$.