Can someone please help me w/ 11b and 12aii.

Ty in advance.

Apologies if this is no longer relevant - missed this the first time.

You know that \(z_3 = \bar{z_2}\) from part i). You can then use the

sum and product of roots formulas to figure out the coefficients from the hints given.

For the second question, try expanding out and equating the real and imaginary parts to zero. It's not the hardest to follow from there.

Hello anyone, I'm having trouble with this question and the thread on it wasn't very helpful. I'm trying to approach it by finding the values of the denominator function, but I've gotten stuck. So far i have that there's obviously an intercept at (-1,0) and the turning point has a y value of -4 because of the range, but i cant think of anything else. any help would be very appreciated!

With this approach, you're effectively solving \(x^2+2ax+b = 0 \implies x = \frac{-2a \pm \sqrt{4a^2-4b}}{2} = -a \pm \sqrt{a^2-b}\). You're told that one of the asymptotes is at -1, so \((-1)^2 -2a+b = 0 \implies b = 2a-1\). Substituting, we then have the asymptotes being at \(-a \pm \sqrt{a^2-(2a - 1)} = -a \pm (a-1)\). Clearly, one of the asymptotes is at -1, as expected and the other one is at \(-2a+1\).

I'm not particularly sure why you think the turning point has a y-value of -4 because of the range. The range and the y-values are almost analogous. The turning point must be at 1/4. Since the function is just the reciprocal of a quadratic multiplied by a constant, the extrema of the quadratic must also be extrema of the reciprocal, so what we are effectively looking for is the x-value of the turning point of the quadratic. Recall that for a quadratic of the form \(ax^2+bx+c\) that the turning point is located at \(x = -\frac{b}{2a}\).

Hopefully that's enough to get you a better understanding and make the problem easier to solve.