Find the modulus and argument of the complex numbers: 1^(°)…
Find the modulus and argument of the complex numbers: $$ \begin{aligned} & 1^{\circ}\; -i; \quad 2^{\circ}\; 3i; \quad 3^{\circ}\; 5; \quad 4^{\circ}\; 1+i; \quad 5^{\circ}\; 1-i; \quad 6^{\circ}\; 2+5i; \quad 7^{\circ}\; 3-4i; \\ & 8^{\circ}\; -3-5i. \end{aligned} $$
Explanation of Method
For $z = x + iy$, the modulus is $|z| = \sqrt{x^2 + y^2}$, and the argument $\arg(z)$ is the angle $\theta$ satisfying $x = |z|\cos\theta$, $y = |z|\sin\theta$, so that $\tan\theta = \dfrac{y}{x}$ (when $x \neq 0$). The quadrant of $(x, y)$ determines which form of $\arctan$ to use:
- Quadrant I ($x>0, y>0$): $\theta = \arctan\!\left(\dfrac{y}{x}\right)$
- Quadrant II ($x<0, y>0$): $\theta = \pi - \arctan\!\left(\dfrac{|y|}{|x|}\right)$
- Quadrant III ($x<0, y<0$): $\theta = -\pi + \arctan\!\left(\dfrac{|y|}{|x|}\right)$
- Quadrant IV ($x>0, y<0$): $\theta = -\arctan\!\left(\dfrac{|y|}{x}\right)$
- Positive real axis ($y=0, x>0$): $\theta = 0$ · Negative real axis ($y=0, x<0$): $\theta = \pi$
- Positive imaginary axis ($x=0, y>0$): $\theta = \dfrac{\pi}{2}$ · Negative imaginary axis ($x=0, y<0$): $\theta = -\dfrac{\pi}{2}$
Step 1° · $z = -i$
$z = 0 + i(-1)$, so $x = 0$, $y = -1$. Modulus: $|z| = \sqrt{0^2 + (-1)^2} = 1$. The point $(0,-1)$ lies on the negative imaginary axis.
$$ \boxed{|z| = 1, \quad \arg(z) = -\frac{\pi}{2}\ (-90°)} $$
Step 2° · $z = 3i$
$z = 0 + i(3)$, so $x = 0$, $y = 3$. Modulus: $|z| = \sqrt{0^2 + 3^2} = 3$. The point $(0,3)$ lies on the positive imaginary axis.
$$ \boxed{|z| = 3, \quad \arg(z) = \frac{\pi}{2}\ (90°)} $$
Step 3° · $z = 5$
$z = 5 + i(0)$, so $x = 5$, $y = 0$. Modulus: $|z| = \sqrt{5^2 + 0^2} = 5$. The point $(5,0)$ lies on the positive real axis.
$$ \boxed{|z| = 5, \quad \arg(z) = 0\ (0°)} $$
Step 4° · $z = 1 + i$
$z = 1 + i(1)$, so $x = 1$, $y = 1$. Modulus: $|z| = \sqrt{1^2 + 1^2} = \sqrt{2}$. Point $(1,1)$ is in Quadrant I, $\tan\theta = 1$.
$$ \boxed{|z| = \sqrt{2}, \quad \arg(z) = \frac{\pi}{4}\ (45°)} $$
Step 5° · $z = 1 - i$
$z = 1 + i(-1)$, so $x = 1$, $y = -1$. Modulus: $|z| = \sqrt{1^2 + (-1)^2} = \sqrt{2}$. Point $(1,-1)$ is in Quadrant IV, $\tan\theta = -1$.
$$ \boxed{|z| = \sqrt{2}, \quad \arg(z) = -\frac{\pi}{4}\ (-45°)} $$
Step 6° · $z = 2 + 5i$
$z = 2 + i(5)$, so $x = 2$, $y = 5$. Modulus: $|z| = \sqrt{2^2 + 5^2} = \sqrt{29}$. Point $(2,5)$ is in Quadrant I.
$$ \boxed{|z| = \sqrt{29}, \quad \arg(z) = \arctan\!\left(\frac{5}{2}\right) \approx 1.190\text{ rad} \approx 68.2°} $$
Step 7° · $z = 3 - 4i$
$z = 3 + i(-4)$, so $x = 3$, $y = -4$. Modulus: $|z| = \sqrt{3^2 + (-4)^2} = \sqrt{25} = 5$. Point $(3,-4)$ is in Quadrant IV.
$$ \boxed{|z| = 5, \quad \arg(z) = -\arctan\!\left(\frac{4}{3}\right) \approx -0.927\text{ rad} \approx -53.1°} $$
Step 8° · $z = -3 - 5i$
$z = -3 + i(-5)$, so $x = -3$, $y = -5$. Modulus: $|z| = \sqrt{(-3)^2 + (-5)^2} = \sqrt{34}$. Point $(-3,-5)$ is in Quadrant III.
$$ \arg(z) = -\pi + \arctan\!\left(\frac{5}{3}\right) \approx -\pi + 1.030 \approx -2.112\text{ rad} \approx -121.0° $$
Equivalently, $\arg(z) = \pi + \arctan\!\left(\dfrac{-5}{-3}\right) = \pi + \arctan\!\left(\dfrac{5}{3}\right)$, or $\approx 4.171$ rad using the $[0, 2\pi)$ convention.
$$ \boxed{|z| = \sqrt{34}, \quad \arg(z) = -\pi + \arctan\!\left(\frac{5}{3}\right)} $$
Summary Table
| No. | Complex number | Modulus | Argument |
|---|---|---|---|
| $1^{\circ}$ | $-i$ | $1$ | $-\frac{\pi}{2}$ |
| $2^{\circ}$ | $3i$ | $3$ | $\frac{\pi}{2}$ |
| $3^{\circ}$ | $5$ | $5$ | $0$ |
| $4^{\circ}$ | $1+i$ | $\sqrt{2}$ | $\frac{\pi}{4}$ |
| $5^{\circ}$ | $1-i$ | $\sqrt{2}$ | $-\frac{\pi}{4}$ |
| $6^{\circ}$ | $2+5i$ | $\sqrt{29}$ | $\arctan(\frac{5}{2})$ |
| $7^{\circ}$ | $3-4i$ | $5$ | $-\arctan(\frac{4}{3})$ |
| $8^{\circ}$ | $-3-5i$ | $\sqrt{34}$ | $-\pi + \arctan(\frac{5}{3})$ |