How to solve that quadratic complex equation? \frac{z^{2}-1}{z^{2}+z+1} So, there can't be

Answered question

2022-01-17

How to solve that quadratic complex equation?
z21z2+z+1
So, there cant

Answer & Explanation

nick1337

nick1337

Expert2022-01-19Added 777 answers

Step 1Writingz=x+iythenz2+z+1=(x+iy)2+(x+iy)+1=(x2y2+x+1i(2xy+y)Step 2If(x2y2+x+1)+i(2xy+y)=0with x,y real numbers, then you needx2y2+x+1=0and2xy+y=0You solve these the way you usually solve equations for real numbers.So, for example, you have0=2xy+y=(2x+1)ySo either y=0 or 2x+1=0. If y=0, then the first equation reduces to x2+x+1=0, which has no real solutions, so there are no solutions with y=0. If y0, then 2x+1=0, so x=12. Plugging into the first equation, we get0=14y212+1=y2+34so you get thaty2=34ory=±32So the two solutions arez=12+i32andz=12i32Step 3The acrobatics from step 2 are unnecessary. You don't have to decompose into real and imaginary parts, because the quadratic formula works for complex numbers! (Provided you take complex square roots). Sincez2+z+1=(z+12)2+34(by completing the square), then this is zero if and only ifz+12=34if and only ifz=12+32You may recognize this as exactly what you get from the quadratic formula applied toz2+z+1and you may also recognize them as the solutions you get if you go through the contorsions of step 2 above. So just find the two complex square roots of -3, and rejoice! (The quadratic formula works even if the coefficients of the quadratic are complex numbers, instead of real numbers).
Vasquez

Vasquez

Expert2022-01-19Added 669 answers

The maximal domain of definition is,as you imply, the complement in C of the set of roots of the polynomial z2+z+1 Therefore to find the bad points, we need to solve the equation z2+z+1=0 I am pretty sure, now, that you know how to solve quadratic equations, no?

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?