Question

question 2
2.1 find all complex numbers x such that $x^{3}+1 = 0$. give the answers in polar form with the angles in radians (8)
2.2 write the polar co - ordinates $(5;\frac{pi}{2})$ in cartesian form. (3)
2.3 solve for x and y if $2x+jy=(1 + j2)(-2 - j3)$ (5)
2.4 name and sketch the curve $y^{2}=49 - x^{2}$. (3)

Explanation:

Step1: Solve $x^{3}+1 = 0$

Rewrite as $x^{3}=-1$. In polar form, $- 1=\cos\pi + i\sin\pi$. By De - Moivre's theorem, $x = r(\cos\theta+i\sin\theta)$ and $x^{3}=r^{3}(\cos3\theta + i\sin3\theta)$. So $r^{3}=1$ (then $r = 1$) and $3\theta=2k\pi+\pi$, $\theta=\frac{2k\pi+\pi}{3}$, $k = 0,1,2$.
When $k = 0$, $\theta=\frac{\pi}{3}$, $x_1=\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}=\frac{1 + i\sqrt{3}}{2}$; when $k = 1$, $\theta=\pi$, $x_2=\cos\pi+i\sin\pi=-1$; when $k = 2$, $\theta=\frac{5\pi}{3}$, $x_3=\cos\frac{5\pi}{3}+i\sin\frac{5\pi}{3}=\frac{1 - i\sqrt{3}}{2}$. In polar form: $x_1=\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}$, $x_2=\cos\pi+i\sin\pi$, $x_3=\cos\frac{5\pi}{3}+i\sin\frac{5\pi}{3}$.

Step2: Convert polar to Cartesian

For polar coordinates $(r,\theta)=(5,\frac{\pi}{2})$, use $x = r\cos\theta$ and $y = r\sin\theta$. So $x=5\cos\frac{\pi}{2}=0$ and $y = 5\sin\frac{\pi}{2}=5$. The Cartesian form is $(0,5)$.

Step3: Expand and solve for $x$ and $y$

First expand $(1 + j2)(-2 - j3)=-2-3j-4j - 6j^{2}=-2-7j + 6=4-7j$. So $2x+jy=4-7j$. Then $2x = 4$ (so $x = 2$) and $y=-7$.

Step4: Identify and sketch the curve

Rewrite $y^{2}=49 - x^{2}$ as $x^{2}+y^{2}=49$. This is a circle with center at the origin $(0,0)$ and radius $r = 7$.

Answer:

2.1: $x_1=\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}$, $x_2=\cos\pi+i\sin\pi$, $x_3=\cos\frac{5\pi}{3}+i\sin\frac{5\pi}{3}$
2.2: $(0,5)$
2.3: $x = 2$, $y=-7$
2.4: Circle with center $(0,0)$ and radius $7$