Question
Consider \(z=\cos \Theta +i\sin \Theta \) where \(z\epsilon C ,z\neq 1.\)
Show that\(Re\left ( \frac{1+z}{1-z} \right )=0\)
Answer/Explanation
$$
\frac{1+z}{1-z}=\frac{1+\cos \theta+\mathrm{i} \sin \theta}{1-\cos \theta-\mathrm{i} \sin \theta}
$$
attempt to use the complex conjugate of their denominator
$$
\begin{aligned}
& =\frac{(1+\cos \theta+\mathrm{i} \sin \theta)(1-\cos \theta+\mathrm{i} \sin \theta)}{(1-\cos \theta-\mathrm{i} \sin \theta)(1-\cos \theta+\mathrm{i} \sin \theta)} \\
& \operatorname{Re}\left(\frac{1+z}{1-z}\right)=\frac{1-\cos ^2 \theta-\sin ^2 \theta}{(1-\cos \theta)^2+\sin ^2 \theta}\left(=\frac{1-\cos ^2 \theta-\sin ^2 \theta}{2-2 \cos \theta}\right)
\end{aligned}
$$
using $\cos ^2 \theta+\sin ^2 \theta=1$ to simplify the numerator
$$
\operatorname{Re}\left(\frac{1+z}{1-z}\right)=0
$$
Question
Find the gradient of the tangent to the curve \({x^3}{y^2} = \cos (\pi y)\) at the point (−1, 1) .
▶️Answer/Explanation
Sol:
METHOD 1
$$
\begin{aligned}
& 3 x^2 y^2+2 x^3 y \frac{\mathrm{d} y}{\mathrm{~d} x}=-\pi \sin (\pi y) \frac{\mathrm{d} y}{\mathrm{~d} x} \\
& \text { At }(-1,1), 3-2 \frac{\mathrm{d} y}{\mathrm{~d} x}=0 \\
& \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{3}{2}
\end{aligned}
$$
METHOD 2
$$
\begin{aligned}
& 3 x^2 y^2+2 x^3 y \frac{\mathrm{d} y}{\mathrm{~d} x}=-\pi \sin (\pi y) \frac{\mathrm{d} y}{\mathrm{~d} x} \\
& \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{3 x^2 y^2}{-\pi \sin (\pi y)-2 x^3 y} \\
& \text { At }(-1,1), \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{3(-1)^2(1)^2}{-\pi \sin (\pi)-2(-1)^3(1)}=\frac{3}{2}
\end{aligned}
$$