I'm gonna need a bit help here fellas.
The equation is
iz^2+(1+i)z-7+4i=0
iz^2+(1+i)z-7+4i=0 - Wolfram|Alpha
Can someone guide me through this one?
I'm gonna need a bit help here fellas.
The equation is
iz^2+(1+i)z-7+4i=0
iz^2+(1+i)z-7+4i=0 - Wolfram|Alpha
Can someone guide me through this one?
Hello, cristianodoni72!
$\displaystyle \text{Solve for }z\!:\;\;iz^2+(1+i)z-(7-4i)\:=\:0$
Quadratic Formula:
. . $\displaystyle z \;=\;\frac{\text{-}(1+i) \pm \sqrt{(1+i)^2 + 4(i)(7-4i)}}{2(i)} \;=\; \frac{\text{-}(1+i)\pm\sqrt{16 + 30i}}{2i} $
It happens that: .$\displaystyle 16 + 30i \:=\:(5+3i)^2$
So we have: .$\displaystyle z \;=\;\frac{\text{-}(1+i) \pm(5+3i)}{2i} $
The two answers are: .$\displaystyle \begin{Bmatrix}\dfrac{\text{-}(1+i) + (5+3i)}{2i} &=&\dfrac{4+2i}{2i} &=& \dfrac{2}{i} + 1 &=& 1 - 2i \\ \\[-3mm] \dfrac{\text{-}(1+i) - (5+3i)}{2i} &=& \dfrac{\text{-}6-4i}{2i} &=& \text{-}\dfrac{3}{i} - 2 &=& \text{-}2 + 3i \end{Bmatrix}$