Question

question 2
2.1 find all complex numbers x such that x³ + 1 = 0. give the answers in polar form with the angles in radians (8)
2.2 write the polar co - ordinates (5;π/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 y² = 49 - x². (3)

Explanation:

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

Rewrite as $x^{3}=-1$. In polar form, $-1=\cos\pi + j\sin\pi$. By De - Moivre's theorem, if $x = r(\cos\theta + j\sin\theta)$, then $x^{n}=r^{n}(\cos(n\theta)+j\sin(n\theta))$. Here $n = 3$, $r^{3}=1$ (so $r = 1$) and $3\theta=(2k + 1)\pi,k\in\mathbb{Z}$. Then $\theta=\frac{(2k + 1)\pi}{3},k = 0,1,2$. When $k = 0,\theta=\frac{\pi}{3},x_1=\cos\frac{\pi}{3}+j\sin\frac{\pi}{3}$; when $k = 1,\theta=\pi,x_2=\cos\pi+j\sin\pi$; when $k = 2,\theta=\frac{5\pi}{3},x_3=\cos\frac{5\pi}{3}+j\sin\frac{5\pi}{3}$.

Step2: Convert polar to Cartesian

For polar coordinates $(r,\theta)=(5,\frac{\pi}{2})$, using $x = r\cos\theta$ and $y = r\sin\theta$. Then $x=5\cos\frac{\pi}{2}=0$ and $y = 5\sin\frac{\pi}{2}=5$. So the Cartesian coordinates are $(0,5)$.

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 $2x+jy=4-7j$. Equating real and imaginary parts, we have $2x=4$ (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. $x_1=\cos\frac{\pi}{3}+j\sin\frac{\pi}{3},x_2=\cos\pi+j\sin\pi,x_3=\cos\frac{5\pi}{3}+j\sin\frac{5\pi}{3}$
2.2. $(0,5)$
2.3. $x = 2,y=-7$
2.4. It is a circle with center $(0,0)$ and radius $7$.