1. ## more complex numbers

hey guys! i have to solve this equation but i have tried diferent things n i cant work it out. please help?

z^2+2z+iz=0

i have tired taking a common factor out but that doesnt really get me anywhrere...

any ideas?

thank you.

2. First thing to note is that it is a quadratic, so you will get 2 solutions (roots) to the equation.

You are correct to factorize the z which is common in each term i.e.

z(z+2+i) = 0

This implies that z = 0 is the first solution.

Then you have z+2+i=0 which when re-arranged gives

z = -2-i

You can check that you are correct by substituting in these values for z into the original quadratic equation - you will find the left hand side equates to 0 for both z's as required.

3. thank you so much! i'm so tired , didnt even realise the answer was right infront of me! thank you!

4. Originally Posted by boomshine57th
hey guys! i have to solve this equation but i have tried diferent things n i cant work it out. please help?

z^2+2z+iz=0

i have tired taking a common factor out but that doesnt really get me anywhrere...

any ideas?

thank you.
Also you could use the quadratic formula

$\displaystyle z=\frac{(-2+i)-\sqrt{(2+i)^2-0}}{2}=-2-i$ or $\displaystyle z=0$

5. Hello, boomshine57th!

Solve: .$\displaystyle z^2+2z+iz\:=\:0$

i have tried taking a common factor out,
but that doesnt really get me anywhrere. . . . . really?
You know the drill ... factor, then set each factor equal to zero, etc. . . .

Factor: .$\displaystyle z(z + 2 + i) \:=\:0$

Then: .$\displaystyle \begin{array}{ccccc}\boxed{z \:=\:0} \\ z+2+i \:=\:0 & \Rightarrow & \boxed{z \:=\:\text{-}2-i} \end{array}$