The argument of a complex number.

Algebra Level 3

Evaluate $\varphi=\operatorname{arg}\left[ \cos\left(\dfrac\pi2\right)e^{i\pi}+i\sin(2i\pi) \right]$ where $\varphi\in(-\pi,\pi]$ .

The answer is 3.1415.

2 solutions

Chew-Seong Cheong
Sep 7, 2018

$\begin{aligned} \varphi & = \arg \left(\cos \left(\frac \pi 2 \right) e^{i\pi} + i \color{#3D99F6} \sin (2i\pi) \right) & \small \color{#3D99F6} \text{By Euler's formula: }e^{i\theta} = \cos \theta + i\sin \theta \\ & = \arg \left( (0) e^{i\pi} + i \cdot \color{#3D99F6} \frac {e^{-2\pi} - e^{2\pi}}{2i} \right) \\ & \approx \arg \left(-267.745 \right) \\ & = \arg \left(267.745e^{i\pi} \right) \\ & = \boxed \pi & \small \color{#3D99F6} \text{for }\varphi \in (-\pi, \pi] \end{aligned}$

Finn Hulse
May 15, 2014

Because $\cos{\dfrac{\pi}{2}}=0$ , the real part evaluates to $0$ . Taking $\arg{(i\sin{(2i\pi)})}$ , we get $\boxed{\pi}$ .

I stopped after calculating the real part, what is arg? @Finn Hulse

Mardokay Mosazghi - 7 years ago

The argument function.

Finn Hulse - 7 years ago

I'd like to add: i sin(2i(pi)) = - sinh(2(pi)) and since -sinh(2pi) is real, so arg = tan^(-1) (0/-sinh(2pi)) = pi = 3.14(approximate).

John Ashley Capellan - 7 years ago

The argument of a complex number.

The answer is 3.1415.

2 solutions

0 pending reports