Circles and a cardioid in the complex plane from modulus–argument relations, with implicit and parametric gradients

This question investigates curves in the complex plane whose points satisfy \(r=p+q\cos\theta\), where \(p\) and \(q\) are real constants.

Here \(r\) is the modulus and \(\theta\) is an argument of a complex number \(z\). Each pair \((r,\theta)\) fixes one point in the complex plane, so a relation between \(r\) and \(\theta\) traces out a curve.

Begin with \(p=3\) and \(q=0\), so that \(r=3\).

1. Let \(z\) be a complex number with \(r=3\). Find every possible value of \(z\) for which
   1. \(\arg z=\pi\);[1]
   2. \(\operatorname{Re}(z)=\operatorname{Im}(z)\);[1]
   3. \(\arg z=\frac{2\pi}{3}\).[1]
2. On an Argand diagram, sketch the curve \(r=3\), marking the points where it crosses the real and imaginary axes.[2]
3. Next take \(p=0\) and \(q=6\), so that \(r=6\cos\theta\), where \(-\frac\pi2\leq\theta\leq\frac\pi2\). The point representing \(z\) lies on this curve.
   1. Write down the largest possible value of \(r\).[1]
   2. Find \(z\) when \(r\) takes this value.[2]
4. By using \(x=r\cos\theta\) and \(y=r\sin\theta\), where \(x=\operatorname{Re}(z)\) and \(y=\operatorname{Im}(z)\), show that every point of the curve \(r=6\cos\theta\) lies on the circle \((x-3)^2+y^2=9\).[4]
5. Consider the circle \((x-3)^2+y^2=9\).
   1. Express \(\frac{\mathrm{d}y}{\mathrm{d}x}\) as a function of \(x\) and \(y\).[3]
   2. Hence or otherwise, at the points where the gradient of the circle is zero, find \(y\).[2]
6. Sketch the circle \(r=6\cos\theta\), where \(-\frac\pi2\leq\theta\leq\frac\pi2\), on an Argand diagram, marking any intercepts with the axes.[1]
7. Finally, take \(p=3\) and \(q=3\). The Argand diagram below shows the curve \(r=3+3\cos\theta\), where \(0\leq\theta<2\pi\). The tangent to the curve is parallel to the imaginary axis at the points \(D\), \(E\) and \(F\), where \(D\) lies in the second quadrant, \(E\) lies in the third quadrant and \(F\) lies on the positive real axis.
   By writing \(x=r\cos\theta\) and \(y=r\sin\theta\), show that the gradient of the curve at the point \((r\cos\theta,r\sin\theta)\) is \(\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{(1+\cos\theta)(1-2\cos\theta)}{\sin\theta(1+2\cos\theta)}\).[5]
8. Hence, find \(\theta\) at the point \(D\).[2]