1. ## Complex Equations

I'm gonna need a bit help here fellas.

The equation is

iz^2+(1+i)z-7+4i=0
iz&#94;2&#43;&#40;1&#43;i&#41;z-7&#43;4i&#61;0 - Wolfram|Alpha

Can someone guide me through this one?

2. ## Re: Complex Equations

You see you have a quadratic (complex) equation in function of $\displaystyle z$, so just solve it like solving an other quadratic equation.

3. ## Re: Complex Equations

Hello, cristianodoni72!

$\displaystyle \text{Solve for }z\!:\;\;iz^2+(1+i)z-(7-4i)\:=\:0$

. . $\displaystyle z \;=\;\frac{\text{-}(1+i) \pm \sqrt{(1+i)^2 + 4(i)(7-4i)}}{2(i)} \;=\; \frac{\text{-}(1+i)\pm\sqrt{16 + 30i}}{2i}$

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: .$\displaystyle \begin{Bmatrix}\dfrac{\text{-}(1+i) + (5+3i)}{2i} &=&\dfrac{4+2i}{2i} &=& \dfrac{2}{i} + 1 &=& 1 - 2i \\ \\[-3mm] \dfrac{\text{-}(1+i) - (5+3i)}{2i} &=& \dfrac{\text{-}6-4i}{2i} &=& \text{-}\dfrac{3}{i} - 2 &=& \text{-}2 + 3i \end{Bmatrix}$

4. ## Re: Complex Equations

Thanks alot! =) Now I get it!

5. ## Re: Complex Equations

Yes, now that Soroban did it for you, I'm sure you "get it". If he had not told you, would you have recognised that $\displaystyle \sqrt{16+ 30i}= \pm (5+ 3i)$?