I have to find $\arg(z)$ of $5e^{-i\frac{3\pi}2}$

Question

Can anyone help me find $\arg(z)$ of $5e^{-i\frac{3\pi}2}$ ?

Thank you

@Caroline Where did you find this question ? I see that you don't have any idea about exponential form of complex numbers. — , Sep 23 '17 at 15:21
Its a school project - so maybe I have learned it, but just can't remember :-) — Line, Sep 23 '17 at 15:28
Then, you may need to google it. Once you know it, the question is pretty much straightforward. — Danilo Gregorin Afonso, Sep 23 '17 at 17:38

score 0 · Answer 1 · answered Sep 23 '17 at 15:20

0

$$5e^{-3i\pi/2}=5\cos\left(-\dfrac{3\pi}2\right)+i\cdot5\sin\left(-\dfrac{3\pi}2\right)=5i\sin\left(-2\pi+\dfrac\pi2\right)=5i$$

Now use the Definition of atan$2(y,x)$

answered Sep 23 '17 at 15:20

lab bhattacharjee

score 0 · Accepted Answer · answered Sep 23 '17 at 17:12

Following GeorgeCoote's comment:

$$z=|z|\exp(i\arg z),$$

in which $0\leq\arg z< 2\pi$.

Your expression is $z=5\exp(-i\frac{3}{2}\pi)=5\exp\left[i(-\frac{3}{2}\pi)\right]=5\exp\left[i(-\frac{3}{2}\pi+2\pi k)\right]$.

You would get $\arg z= -3/2\pi +2\pi k$ for $k=1$ the value is between $0$ and $2\pi$ hence $\arg z = -3/2\pi +2\pi=\pi/2$.

2 Answers2