Question

question 2
(a) express the complex number z = 3|45° in exponential form.
(b) if $z_1 = 5(cos\frac{pi}{2}+jsin\frac{pi}{2})$ and $z_2 = 4(cos\frac{pi}{3}+jsin\frac{pi}{3})$. find $\frac{z_1}{z_2}$. leave the answer in polar form.
(c) find the square roots of the complex number x if x = - 3 - 4j.
(d) write down the radius and the co - ordinates of the center of the circle $(x - 4)^2+(y + 3)^2 = 16$
(e) write down the equation of the ellipse with center (0;0), one focal point at (0;4) and one vertex at (0;5). 20

Explanation:

Step1: Recall Euler's formula

Euler's formula is $e^{j\theta}=\cos\theta + j\sin\theta$. For a complex number in polar form $z = r\angle\theta$, it can be written as $z = re^{j\theta}$. Given $z = 3\angle45^{\circ}$, and $45^{\circ}=\frac{\pi}{4}$ radians. So $z = 3e^{j\frac{\pi}{4}}$.

Step2: Divide complex - numbers in polar form

If $Z_1=r_1(\cos\theta_1 + j\sin\theta_1)$ and $Z_2=r_2(\cos\theta_2 + j\sin\theta_2)$, then $\frac{Z_1}{Z_2}=\frac{r_1}{r_2}[\cos(\theta_1-\theta_2)+j\sin(\theta_1 - \theta_2)]$. Here, $r_1 = 5$, $\theta_1=\frac{\pi}{2}$, $r_2 = 4$, $\theta_2=\frac{\pi}{3}$. So $\frac{Z_1}{Z_2}=\frac{5}{4}[\cos(\frac{\pi}{2}-\frac{\pi}{3})+j\sin(\frac{\pi}{2}-\frac{\pi}{3})]=\frac{5}{4}(\cos\frac{\pi}{6}+j\sin\frac{\pi}{6})$.

Step3: Find square - roots of a complex number

Let $x=-3 - 4j$. First, find the modulus $r=\sqrt{(-3)^2+(-4)^2}=5$ and $\theta=\arctan(\frac{-4}{-3})-\pi=\arctan(\frac{4}{3})-\pi$ (since the complex number is in the third quadrant). The square - roots of a complex number $z = r(\cos\theta + j\sin\theta)$ are given by $\sqrt{z}=\sqrt{r}(\cos\frac{\theta + 2k\pi}{2}+j\sin\frac{\theta + 2k\pi}{2})$, $k = 0,1$.
For $k = 0$:
$r_0=\sqrt{5}$, $\theta_0=\frac{\arctan(\frac{4}{3})-\pi}{2}$.
For $k = 1$:
$r_1=\sqrt{5}$, $\theta_1=\frac{\arctan(\frac{4}{3})-\pi + 2\pi}{2}=\frac{\arctan(\frac{4}{3})+\pi}{2}$.
Another way: Let $\sqrt{-3 - 4j}=a+bj$. Then $(a + bj)^2=-3 - 4j$, so $a^{2}-b^{2}=-3$ and $2ab=-4$. Solving the system

$$\begin{cases}a^{2}-b^{2}=-3\\ab=-2\end{cases}$$

, from $b =-\frac{2}{a}$, substitute into $a^{2}-b^{2}=-3$: $a^{2}-\frac{4}{a^{2}}=-3$. Let $t = a^{2}$, then $t-\frac{4}{t}=-3$, $t^{2}+3t - 4 = 0$, $(t + 4)(t - 1)=0$. Since $t=a^{2}\geq0$, $t = 1$, $a=\pm1$. If $a = 1$, $b=-2$; if $a=-1$, $b = 2$. So the square - roots are $1-2j$ and $-1 + 2j$.

Step4: Identify center and radius of a circle

The standard form of a circle is $(x - h)^2+(y - k)^2=r^2$, where $(h,k)$ is the center and $r$ is the radius. For the circle $(x - 4)^2+(y + 3)^2=16$, the center is $(4,-3)$ and the radius $r = 4$.

Step5: Find the equation of an ellipse

The standard form of an ellipse centered at the origin is $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$ for a vertical ellipse ($a>b>0$). The distance from the center to a focus is $c$ and the distance from the center to a vertex is $a$. Given $c = 4$ and $a = 5$. Using the relationship $c^{2}=a^{2}-b^{2}$, we have $b^{2}=a^{2}-c^{2}=25 - 16 = 9$. So the equation of the ellipse is $\frac{x^{2}}{9}+\frac{y^{2}}{25}=1$.

Answer:

(a) $3e^{j\frac{\pi}{4}}$
(b) $\frac{5}{4}(\cos\frac{\pi}{6}+j\sin\frac{\pi}{6})$
(c) $1-2j$ and $-1 + 2j$
(d) Center: $(4,-3)$, Radius: $4$
(e) $\frac{x^{2}}{9}+\frac{y^{2}}{25}=1$