This question asks you to explore curves in the complex plane defined by $r=a+b\sin\theta$, where&

Question

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 $r=4$. Suppose that $z$ is a complex number such that $r=4$. Find all possible values of $z$ such that[3] $z$ is purely real; $z$ is purely imaginary; the argument of $z$ is $\frac\pi4$. 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] 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.[3] Write down the greatest possible value of $r$. For this value of $r$, find $z$. 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] Consider the curve $x^2+(y-2)^2=4$.[5] Find an expression for $\frac{\mathrm{d}y}{\mathrm{d}x}$ in terms of $x$ and $y$. Hence or otherwise, find the values of $x$ for which the gradient of the curve is undefined. 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] 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] Hence, find the value of $\theta$ which corresponds to point $A$.[2]

Accepted Answer

(a) $\pm4,\ \pm4i,\ 2\sqrt2+2\sqrt2i$ (c) $4,4i$ (e) $y\prime=-x/(y-2),\ x=\pm2$ (h) $7\pi/6$

Solution:

Typed Solution