Find argument of a complex number

[Aion]

Hi all, I’m trying to understand a solution involving complex numbers in trigonometric form. The problem is:

Find the argument of the complex number $z_1=z^2-z$ if $z=cos(\varphi)+isin(\varphi)$.

The solution proceeds as follows:

$$z_1=\left( cos(\varphi)+isin(\varphi)\right)^2-(cos(\varphi)+isin(\varphi))$$$$z_1=\left( cos(2\varphi)-cos(\varphi)\right)+i\left( sin(2\varphi)-sin(\varphi)\right)$$$$z_1=2sin\left(\frac{\varphi}{2}\right)\left( -sin\left(\frac{3\varphi}{2} \right)+icos\left( \frac{3\varphi}{2}\right)\right)$$
Thus the modulus is

$$\left| z_1\right|=\sqrt{4sin^2\left(\frac{\varphi}{2}\right)\left(sin^2\left( \frac{3\varphi}{2}\right) + cos^2\left( \frac{3\varphi}{2}\right)\right)}=2 \left| sin\left(\frac{\varphi}{2}\right)\right|$$
In accordance with the definition of absolute value, we have to consider three cases:

(a) If $sin\left(\frac{\varphi}{2} \right)=0$, that is $\varphi=2k\pi$, $k$ is any integer, the argument of $z_1$ is not defined.

(b) If $sin\left(\frac{\varphi}{2} \right)>0$, which occurs when $4\pi k<\varphi<(4k+2)\pi$ then $\left| z_1\right|=2sin\left(\frac{\varphi}{2} \right)$ and the trigonometric form of the complex number $z_1$ is as follows:

$$z_1=2sin\left(\frac{\varphi}{2}\right)\left( -sin\left(\frac{3\varphi}{2} \right)+icos\left( \frac{3\varphi}{2}\right)\right)=2sin\left(\frac{\varphi}{2}\right)\left( cos\left(\frac{\pi+3\varphi}{2} \right)+isin\left( \frac{\pi+3\varphi}{2}\right)\right)$$
Consequently, if $\varphi$ satisfy the above condition, then $$arg(z_1)=\frac{\pi+3\varphi}{2}$$
(c) If $sin\left(\frac{\varphi}{2} \right)<0$, that is $(4k+2)\pi<\varphi<(4k+4)\pi$, $k$ any integer then $\left| z_1\right|=-2sin(\frac{\varphi}{2})$

$$z_1=2sin\left(\frac{\varphi}{2}\right)\left( -sin\left(\frac{3\varphi}{2} \right)+icos\left( \frac{3\varphi}{2}\right)\right)=-2sin\left(\frac{\varphi}{2}\right)\left( sin\left(\frac{3\varphi}{2} \right)-icos\left( \frac{3\varphi}{2}\right)\right)=-2sin\left(\frac{\varphi}{2}\right)\left( cos\left(\frac{3\pi+3\varphi}{2} \right)+isin\left( \frac{3\pi+3\varphi}{2}\right)\right)$$
Hence, if $\varphi$ satisfies the condition above, then $$arg(z_1)=\frac{3\pi+3\varphi}{2}$$
I understand most of this derivation, but I’m having trouble fully grasping the step where the expression inside the parentheses is rewritten using the angle identities. Is there an easy way to see that
$$sin\left(\frac{3\varphi}{2} \right)-icos\left( \frac{3\varphi}{2}\right)= cos\left(\frac{3\pi+3\varphi}{2} \right)+isin\left( \frac{3\pi+3\varphi}{2}\right)$$I can see that the result checks out using the addition formulas, but I don’t quite understand how to derive it on my own from the beginning.

 

[Dr.Peterson]

Is there an easy way to see that
$$sin\left(\frac{3\varphi}{2} \right)-icos\left( \frac{3\varphi}{2}\right)= cos\left(\frac{3\pi+3\varphi}{2} \right)+isin\left( \frac{3\pi+3\varphi}{2}\right)$$I can see that the result checks out using the addition formulas, but I don’t quite understand how to derive it on my own from the beginning.

First, I assume you see WHY they want to do this: to put it in the form cos( ) + i sin( ), so they can read off the argument from the expression. In particular, they want that angle to be within whatever range they consider appropriate for the argument (presumably $(-\pi,\pi]$).

So you want to rewrite $\sin\left(\frac{3\varphi}{2} \right)$ as a cosine.

The natural way to do this is the cofunction identity $\sin(\theta)=\cos(\frac{\pi}{2}-\theta)$. Try doing that, and then manipulate it further to get the argument in range. (You may find it helpful to use the fact that sine is an odd function, or that cosine is an even function, to get rid of a subtraction.)

Then, of course, repeat the process with the imaginary part.

 

[Aion]

First, I assume you see WHY they want to do this: to put it in the form cos( ) + i sin( ), so they can read off the argument from the expression. In particular, they want that angle to be within whatever range they consider appropriate for the argument (presumably $(-\pi,\pi]$).

So you want to rewrite $\sin\left(\frac{3\varphi}{2} \right)$ as a cosine.

The natural way to do this is the cofunction identity $\sin(\theta)=\cos(\frac{\pi}{2}-\theta)$. Try doing that, and then manipulate it further to get the argument in range. (You may find it helpful to use the fact that sine is an odd function, or that cosine is an even function, to get rid of a subtraction.)

Then, of course, repeat the process with the imaginary part.

We have one system of equations to solve.

1. $cos(x)=sin\left(\frac{3\varphi}{2}\right)$
2. $sin(x)=-cos\left(\frac{3\varphi}{2}\right)$

We first solve equation 1.
$$cos(x)=sin\left(\frac{3\varphi}{2}\right)=cos\left(\frac{\pi}{2}-\frac{3\varphi}{2} \right)$$$$x=\pm \left( \frac{\pi}{2}-\frac{3\varphi}{2}\right)+2\pi k$$$$x_1=\frac{\pi}{2}-\frac{3\varphi}{2}+2\pi k$$$$x_2=\frac{3\varphi}{2}-\frac{\pi}{2}+2\pi k$$
Then we solve equation 2.
$$sin(x)=-cos\left(\frac{3\varphi}{2}\right)=-sin\left(\frac{\pi}{2}-\frac{3\varphi}{2} \right)=sin\left(\frac{3\varphi}{2}-\frac{\pi}{2} \right)$$$$x_1=\frac{3\varphi}{2}-\frac{\pi}{2}+2\pi k$$$$x_2=\frac{3\pi}{2}-\frac{3\varphi}{2}+2\pi k$$
Comparing the two solutions for $x$. One value appears in both equations:
$$x=\frac{3\varphi}{2}-\frac{\pi}{2}+2\pi k$$If we let $k=1$, then we have
$$x=\frac{3\varphi+3\pi}{2}$$

 

[blamocur]

I am too lazy to draw a diagram, but here is a geometric approach:

1. $|z| = |z^2| = 1$, which means that 0, $z$ and $z^2$ form an isoscless triangle.
2. In such triangles the base, i.e. $z^2-z$, is orthogonal to the bisector of the apex angle,
3. The argument of the bisector is $\frac{1}{2}(\phi+2\phi) = \frac{3\phi}{2}$
4. The argument of base is thus $\frac{3\phi}{2}+\frac{\pi}{2}$.

The reason I like this geometric approach is that it works for arbitrary powers $z^n - z^m$.

 

[logistic_guy]

Find the argument of the complex number $z_1=z^2-z$ if $z=cos(\varphi)+isin(\varphi)$.

I like to solve this problem by using polar form of Euler's formula and the property of multiplying two arguments.

$\displaystyle \text{Polar Euler's formula:}$
$\displaystyle e^{i\varphi} = \cos \varphi + i\sin \varphi$

$\displaystyle \text{The argument property of a product:}$
$\displaystyle \text{arg}(z_1z_2) = \text{arg}(z_1) + \text{arg}(z_2)$


We have $\displaystyle \large z_1 = z^2 - z = z(z - 1) = e^{i\varphi}(e^{i\varphi} - 1)$

Then,

$\displaystyle \large \text{arg}(z_1) = \text{arg}(e^{i\varphi}(e^{i\varphi} - 1)) = \text{arg}(e^{i\varphi}) + \text{arg}{(e^{i\varphi} - 1})$

Since $\displaystyle \large e^{i\varphi}$ is a vector starts at the origin and ends on a point on the unit circle, its argument is:
$\displaystyle \large \text{arg}(e^{i\varphi}) = \varphi$

Since $\displaystyle \large e^{i\varphi} - 1$ is a vector starts at the point $\displaystyle \large (1, 0)$ on the unit circle and ends on the point $\displaystyle \large (\cos \varphi, \sin \varphi)$ on the unit circle, we can draw two lines from the origin of the unit circle to connect the tail and the head of that vector. This forms an isosceles triangle.

The angle bisector of this isosceles triangle with the chord of that vector is $\displaystyle \large \frac{\varphi}{2}$. If we rotate this bisector $\displaystyle 90$ degrees $\displaystyle \large \left(\frac{\varphi}{2} + \frac{\pi}{2}\right)$, we get the direction of the vector $\displaystyle \large e^{i\varphi} - 1$.

Then,

$\displaystyle \large \text{arg}(e^{i\varphi} - 1) = \frac{\varphi}{2} + \frac{\pi}{2}$

Finally, we get:

$\displaystyle \large \text{arg}(z_1) = \text{arg}(e^{i\varphi}) + \text{arg}{(e^{i\varphi} - 1}) = \varphi + \frac{\varphi}{2} + \frac{\pi}{2} = \frac{3\varphi}{2} + \frac{\pi}{2}$

 

[Aion]

I am too lazy to draw a diagram, but here is a geometric approach:

1. $|z| = |z^2| = 1$, which means that 0, $z$ and $z^2$ form an isoscless triangle.
2. In such triangles the base, i.e. $z^2-z$, is orthogonal to the bisector of the apex angle,
3. The argument of the bisector is $\frac{1}{2}(\phi+2\phi) = \frac{3\phi}{2}$
4. The argument of base is thus $\frac{3\phi}{2}+\frac{\pi}{2}$.

The reason I like this geometric approach is that it works for arbitrary powers $z^n - z^m$.

Interesting metod. Can we obtain both solutions with this approach, or does it assume that $0<\varphi<\pi$? I will attempt to draw a diagram of your method.

 

[blamocur]

Can we obtain both solutions with this approach, or does it assume that 0<φ<π0<\varphi<\pi0<φ<π?

Good point. It becomes $\frac{3\phi-\pi}{2}$ for $\pi < \phi < 2\pi$. Here is an "algebra-geometric" reasoning:
The angle between two complex numbers $u$ and $v$ measured from $u$ to $v$ is the argument of $\frac{v}{u}$
Bisector's direction is given by $z^2 + z$, thus we are interested in the argument of $\frac{z^2-z}{z^2+z} = \frac{z-1}{z+1}$.
But:
$$\frac{z-1}{z+1} = \frac{(z-1)(\bar z +1)}{(z+1)(\bar z+1)} = \frac{|z|^2-1 + 2\Im (z)}{|z+1|^2} = \Im(z)\frac{2}{|z+1|^2}$$Thus the resulting angle is that of $\Im(z)$, i.e. the imaginary part of $z$, which is $\pm\frac{\pi}{2}$ depending on the argument of $z$.

 

[blamocur]

There is some ambiguity here. Consider $\phi_1 = -\frac{\pi}{2}$ and $\phi_2 = \frac{3\pi}{2}$. For both cases $z = -i$ and $\theta = \arg(z^2-z) = \arg(i-1) = \frac{3\pi}{4}$. But
$$\theta - \frac{3\phi_1}{2} = \frac{3\pi}{2}\equiv -\frac{\pi}{2} \;\;\;\;\;\text{and}\;\;\; \theta - \frac{3\phi_1}{2} = -\frac{3\pi}{2} \equiv\frac{\pi}{2}$$I.e., if $\phi$ is defined $\mod 2\pi$ then $\frac{3\phi}{2}$ is defined $\mod \pi$