Finding the complex roots of an equation: z^2 - 4 cos(alpha) + 4 = 0, 0<alpha<pi/2

[tpupble]

I attempted to use the quadratic formula to solve the roots using a quadratic of the form ax²+bx+c, with Z replacing x. Why does the answer use -4cos(a) as the b term in this general equation when there is no Z in front of it? Shouldn’t the b term be equal to 0 and the c term equal to 4-4cos(a)?

 

[Dr.Peterson]

View attachment 36547I attempted to use the quadratic formula to solve the roots using a quadratic of the form ax²+bx+c, with Z replacing x. Why does the answer use -4cos(a) as the b term in this general equation when there is no Z in front of it? Shouldn’t the b term be equal to 0 and the c term equal to 4-4cos(a)?

Clearly it's a typo; they left out the z in the middle term of the problem.

 

[pka]

Write it as:
[imath]z^2-4\cos(\alpha)+4=0\\ z^2=4\cos(\alpha)-4\\z=\pm\sqrt{4\cos(\alpha)-4}[/imath]