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




2022-01-19Added 573 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).



2022-01-19Added 457 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?

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