Help me this one complex number question (1 Viewer)

upishcat · Dec 8, 2019

Screen Shot 2019-12-08 at 8.26.07 pm.png

I turned z1 and z2 into modulus argument form so I got z1 = cos(pi/2)+isin(pi/2) and z2= cos(pi/4)+isin(pi/4) but I dont know how to prove arg(z1+z2) as I had never seen this identity before.

Drdusk · Dec 8, 2019

upishcat said:
View attachment 27898
I turned z1 and z2 into modulus argument form so I got z1 = cos(pi/2)+isin(pi/2) and z2= cos(pi/4)+isin(pi/4) but I dont know how to prove arg(z1+z2) as I had never seen this identity before.

Why not just add the two complex numbers as they are? This gives you z1 + z2 = 'something', and then just find the argument using

$\arg(z_1 + z_2) = \tan^{-1}\bigg({\dfrac{\Im(z_1+z_2)}{\Re(z_1+z_2)}}\bigg)$

Argument is calculated by taking tan inverse of the imaginary part over the real part of a complex number.

fan96 · Dec 8, 2019

Drdusk said:
Why not just add the two complex numbers as they are? This gives you z1 + z2 = 'something', and then just find the argument using

$\arg(z_1 + z_2) = \tan^{-1}\bigg({\dfrac{\Im(z_1+z_2)}{\Re(z_1+z_2)}}\bigg)$

Argument is calculated by taking tan inverse of the imaginary part over the real part of a complex number.

The argument of a complex number satisfies

$\tan(\arg z) = \frac{\text{Im } z}{\text{Re } z}.$

This is not the same as

$\arg z = \tan^{-1}\left(\frac{\text{Im } z}{\text{Re } z}\right)$

because $\tan^{-1}$ is not an inverse function of $\tan$ .

With $z = \exp({3\pi}/4 \cdot i)$ it's clear that the above does not hold, as the function $\tan^{-1} : \mathbb R \to (-\pi/2, \pi/2)$ can't possibly have an output of $3\pi/4$ .

In the case of this question, both numbers have modulus 1. If you draw a diagram you can probably see that the argument of the sum will be the average of the sum of the arguments ( $z_1$ , $z_2$ and the origin form an isosceles triangle).

Drdusk · Dec 8, 2019

fan96 said:
The argument of a complex number satisfies

$\tan(\arg z) = \frac{\text{Im } z}{\text{Re } z}.$

This is not the same as

$\arg z = \tan^{-1}\left(\frac{\text{Im } z}{\text{Re } z}\right)$

because $\tan^{-1}$ is not an inverse function of $\tan$ .

With $z = \exp({3\pi}/4 \cdot i)$ it's clear that the above does not hold, as the function $\tan^{-1} : \mathbb R \to (-\pi/2, \pi/2)$ can't possibly have an output of $3\pi/4$ .

In the case of this question, both numbers have modulus 1. If you draw a diagram you can probably see that the argument of the sum will be the average of the sum of the arguments.

I waaas waiting for someone to say that imao.

It's such a small subtlety kinda negligible.

fan96 · Dec 8, 2019

Drdusk said:
It's such a small subtlety kinda negligible.

It's not negligible at all - it's the biggest defect in the "inverse" trig functions and it can completely throw off your calculations if you're not careful.

If instead we had $z_2 = -1$ then this method would've given a very wrong answer.

Suppose you're developing a navigation software with some sort of compass function.
Ignoring the fact that we don't actually live in two-dimensional space, if you just use the method discussed above, a bearing of 315 degrees is calculated as 135 degrees and now some poor bushwalker is probably very badly lost.

upishcat · Dec 8, 2019

Thank you very much for the help guys.

HeroWise · Dec 9, 2019

You can do it geometrically too, Since its a rhobus yada yada

Help me this one complex number question (1 Viewer)

upishcat

Member

Drdusk

Moderator

fan96

617 pages

Drdusk

Moderator

fan96

617 pages

upishcat

Member

HeroWise

Active Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)