What is the argument of 8i?

Question

Since 8i don't have any real part. so, we can assume it's real part as 0 but tangent of zero degree is undefined. Then what is the argument of it?

Answer 1

By definition the principal argument $\operatorname{Arg}(z)$ in usually defined in the interval $(-\pi,\pi]$ therefore

$$\operatorname{Arg}(8i)=\pi/2$$

If we do not refer to the principal value therefore we can use any

$$\arg(8i)=\pi/2+2k\pi \quad k\in \mathbb{Z}$$

that is

$$8i=8e^{i(\pi/2+2k\pi)}$$

Answer 2

It's $\dfrac{\pi}{2}$. Just find the point $8i$ in the complex plane.

Answer 3

In the complex plane, $8i$ is on the positive part of the $y-$axis, so the angle from the positive part of the $x-$axis is well-defined and equal to $\frac{\pi}{2}$.

Note that you have : $$8i = 8 e^{i\frac{\pi}{2}}$$

Answer 4

Yes, $8i=0+8i$ is on the imaginary axis. The argument is $\frac \pi 2$ as that is the angle from the positive real axis to the positive imaginary axis. The formula for the argument uses the arctangent (not the tangent) and breaks down for angles $\pm \frac \pi 2$ because the ratio of imaginary part to real part is not defined.