How to solve that quadratic complex equation? z2−1z2+z+1 So, there cant
How to solve that quadratic complex equation?
So, there cant
Answer & Explanation
Step 1WritingthenStep 2Ifwith x,y real numbers, then you needandYou solve these the way you usually solve equations for real numbers.So, for example, you haveSo either or . If , then the first equation reduces to , which has no real solutions, so there are no solutions with . If , then , so . Plugging into the first equation, we getso you get thatorSo the two solutions areandStep 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). Since(by completing the square), then this is zero if and only ifif and only ifYou may recognize this as exactly what you get from the quadratic formula applied toand 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).
The maximal domain of definition is,as you imply, the complement in of the set of roots of the polynomial
Therefore to find the bad points, we need to solve the equation
I am pretty sure, now, that you know how to solve quadratic equations, no?