Let $z=x+y\mathbf{i}$ Modulus of $z$, denoted by $\left| z \right|$, i.e. \(r=\left| z \r

Question

Let $z=x+y\mathbf{i}$ Modulus of $z$, denoted by $\left| z \right|$, i.e. $r=\left| z \right|=\left| x+y\mathbf{i} \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}$ Argument of $z$, denoted by $\theta $ is the angle made with the positive $x$-axis. Note that $-\pi <\theta \le \pi $ (principal argument) $$\theta = \arg(z) = \arg(x + yi) = \begin{cases} \tan^{-1}|y / x|, & \text{if}\hspace{0.5em} z \text{ is in Quadrant}\hspace{0.5em} 1 \ \pi - \tan^{-1}|y / x|, & \text{if}\hspace{0.5em} z \text{ is in Quadrant}\hspace{0.5em} 2 \ \tan^{-1}|y / x| - \pi, & \text{if}\hspace{0.5em} z \text{ is in Quadrant}\hspace{0.5em} 3 \ -\tan^{-1}|y / x|, & \text{if}\hspace{0.5em} z \text{ is in Quadrant}\hspace{0.5em} 4 \end{cases}$$ Specially, $$\theta = \arg(z) = \begin{cases} 0, & z \text{ is on the positive real axis} \ \frac{\pi}{2}, & \text{if}\hspace{0.5em} z \text{ is on the positive imaginary axis} \ \pi, & \text{if}\hspace{0.5em} z \text{ is on the negative real axis} \ -\frac{\pi}{2}, & \text{if}\hspace{0.5em} z \text{ is on the negative imaginary axis} \ \text{undefined}, & \text{if}\hspace{0.5em} z = 0 \end{cases}$$

Accepted Answer

$|z|=\sqrt{x^2+y^2}$ and the principal argument $\arg z\in(-\pi,\pi]$ is measured anticlockwise from the positive real axis; $\arg 0$ is undefined.

Location	Principal argument
Positive real axis	\(0\)
Positive imaginary axis	\(\pi/2\)
Negative real axis	\(\pi\)
Negative imaginary axis	\(-\pi/2\)
Origin	Undefined

Argand Diagrams, Modulus and Argument

Argand Diagrams, Modulus and Argument

Video Solution:

Solution:

Solution locked