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.
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.
Hello, boomshine57th!
You know the drill ... factor, then set each factor equal to zero, etc. . . .Solve: .$\displaystyle z^2+2z+iz\:=\:0$
i have tried taking a common factor out,
but that doesnt really get me anywhrere. . . . . really?
Factor: .$\displaystyle z(z + 2 + i) \:=\:0$
Then: .$\displaystyle \begin{array}{ccccc}\boxed{z \:=\:0} \\ z+2+i \:=\:0 & \Rightarrow & \boxed{z \:=\:\text{-}2-i} \end{array}$