# Help me this one complex number question (1 Viewer)

#### upishcat

##### New Member

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

##### π
Moderator
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

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

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

#### fan96

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

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

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

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

This is not the same as

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

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

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

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 ($\bg_white z_1$, $\bg_white z_2$ and the origin form an isosceles triangle).

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#### Drdusk

##### π
Moderator
The argument of a complex number satisfies

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

This is not the same as

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

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

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

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

##### 617 pages
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 $\bg_white z_2 = -1$ 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

##### New Member
Thank you very much for the help guys.

#### HeroWise

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