IB AA HL May 2026 P3 TZA Q2

This question asks you to explore curves in the complex plane defined by \(r=a+b\sin\theta\), where \(a\) and \(b\) are real constants.

A complex number \(z\) can be represented by a point in the complex plane whose position is determined by the modulus, \(r\), and the argument, \(\theta\), of \(z\). The set of points which satisfy a given relationship between \(r\) and \(\theta\) form a curve in the complex plane.

First consider \(a=4\) and \(b=0\), so that \(r=4\).

1. Suppose that \(z\) is a complex number such that \(r=4\). Find all possible values of \(z\) such that
   1. \(z\) is purely real;[1]
   2. \(z\) is purely imaginary;[1]
   3. the argument of \(z\) is \(\frac\pi4\).[1]
2. Sketch the curve defined by \(r=4\) on an Argand diagram, showing clearly the values of the intercepts with the real and imaginary axes.[2]
3. Now consider \(a=0\) and \(b=4\), so that \(r=4\sin\theta\), where \(0\leq\theta\leq\pi\). Suppose that \(z\) is a complex number on this curve.
   1. Write down the greatest possible value of \(r\).[1]
   2. For this value of \(r\), find \(z\).[2]
4. Use the identities \(x=r\cos\theta\) and \(y=r\sin\theta\) to show that the points on the curve defined by \(r=4\sin\theta\) lie on the curve \(x^2+(y-2)^2=4\), where \(x=\operatorname{Re}(z)\) and \(y=\operatorname{Im}(z)\).[4]
5. Consider the curve \(x^2+(y-2)^2=4\).
   1. Find an expression for \(\frac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(x\) and \(y\).[3]
   2. Hence or otherwise, find the values of \(x\) for which the gradient of the curve is undefined.[2]
6. Sketch the circle defined by \(r=4\sin\theta\), where \(0\leq\theta\leq\pi\), on an Argand diagram, showing clearly the values of any intercepts with the real and imaginary axes.[1]
7. Finally, consider \(a=4\) and \(b=4\). The curve \(r=4+4\sin\theta\), where \(0\leq\theta\leq2\pi\), is shown on the following Argand diagram. The curve has a horizontal tangent at points \(A\), \(B\) and \(C\). Their real parts are negative, positive and zero, respectively.
   Use the identities \(x=r\cos\theta\) and \(y=r\sin\theta\) to show that the gradient at \((r\cos\theta,r\sin\theta)\) is \(\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\cos\theta(1+2\sin\theta)}{1-\sin\theta-2\sin^2\theta}\).[5]
8. Hence, find the value of \(\theta\) which corresponds to point \(A\).[2]