# finding the argument of $1-\cos 2\theta -i\sin 2 \theta$

"Let z = $$1-\cos 2\theta -i\sin 2 \theta$$, $$0\leq \theta \leq \pi$$. Find the modulus and argument of z in terms of $$\theta$$ in their simplest forms."

The modulus is pretty easy, I got $$2\sin \theta$$ which is correct.

The problem is, when finding the argument of z, I got $$\tan \alpha = \frac{-1}{\tan \theta}$$, where $$\alpha$$ is the argument.

the correct answer is $$\alpha = \theta - \pi/2$$, however I am curious why $$\alpha = \theta + \pi/2$$ doesn't work. This also fits the equation above, but is wrong. Can someone explain?

• Consider separately the cases of acute and obtuse $\theta$, and bear in mind the argument has to be from $-\pi$ to $\pi$. Alternatively, work out $\cos\alpha,\,\sin\alpha$ for each option (note these change sign of you increase $\alpha$ by $\pi$).
– J.G.
Commented Oct 23, 2021 at 8:42
• Note that you introduced an erroneous solution by going to $\tan\alpha$ (minimal period $\pi$) instead of the full $(\cos\alpha,\sin\alpha)$ (minimal period $2\pi$). You need to go back and check for the correct $\alpha$ modulo $2\pi$. Commented Oct 23, 2021 at 8:46

When you calculate $$\tan \alpha$$, you are specifying the slope of the line connecting the origin to your point $$(1-\cos2\theta,-\sin2\theta)$$. What's missing is the direction traveled along this line with the origin as the starting point.

Your two proposed arguments separated by a difference of $$\pi$$, represent the two opposite directions possible on your line, only one of which is correct for actually getting from the origin to your destination. To orient yourself, multiply the cosine of each possible argument by the modulus you found, $$2\sin\theta$$, and compare with the intended $$x$$-coordinate $$1-\cos2\theta$$. Only the candidate $$\alpha=\theta-\pi/2$$ points you the right way.

• Shouldn't the point be $(1-\cos 2 \theta, -\sin 2 \theta)$? If so, this explains it.
– Ray
Commented Oct 23, 2021 at 19:51
• Corrected. Mistyped. Thank you. Commented Oct 24, 2021 at 16:46

$$z = 1-\cos 2\theta -i\sin 2 \theta$$

Modulus

$$|z| = \sqrt{(1-\cos 2 \theta)^2 + (-\sin 2 \theta)^2} = \\ = \sqrt{1 + \cos^2 2 \theta - 2 \cos 2 \theta + \sin^2 2 \theta} = \\ = \sqrt{2 - 2 \cos 2 \theta} = \\ = \sqrt{2 - 2 \cos^2 \theta + 2 \sin^2 \theta} = \\ = \sqrt{2 - 2 \cos^2 \theta + 2 (1-\cos^2 \theta)} = \\ = \sqrt{2 - 2 \cos^2 \theta + 2 -2\cos^2 \theta} = \\ = \sqrt{4 - 4 \cos^2 \theta} = \\ = 2 \sqrt{1 - \cos^2 \theta} = \\ 2 |\sin\theta|.$$

Argument

First of all, notice that $$\eta = \arctan\frac{- \sin 2\theta}{1 - \cos 2\theta} =\arctan\frac{- 2 \sin \theta \cos \theta}{1 - \cos^2 \theta + \sin^2 \theta} =\arctan\frac{- 2 \sin \theta \cos \theta}{2\sin^2 \theta} = \\ = -\arctan\text{cotan} \theta = \theta - \frac{\pi}{2}.$$

since:

$$\arctan(\text{cotan} \beta) = \frac{\pi}{2} - \beta.$$

$$\mathcal{Re}[z] = 1-\cos 2 \theta \geq 0 ~\forall \theta \in [0, \pi]$$

Then, in this case, under the assumption that $$\arg{z} \in (-\pi, \pi]$$:

$$\arg{z} = \eta = \theta - \frac{\pi}{2}.$$

Why does $$\alpha = \theta + \frac{\pi}{2}$$ not work?

First of all, notice that:

$$\mathcal{Im}[z] = -\sin 2 \theta \geq 0 ~\forall \theta \in \left[\frac{\pi}{2}, \pi\right]$$

Hence, for $$\theta \in \left[\frac{\pi}{2}, \pi\right]$$, $$z$$ represents a vector in the first quadrant of the complex plane. This means that its argument must be positive and less that $$\frac{\pi}{2}$$. Indeed:

$$\eta = \theta - \frac{\pi}{2} \in \left[0, \frac{\pi}{2}\right] ~\forall \theta \in \left[\frac{\pi}{2}, \pi\right],$$

while

$$\alpha = \theta + \frac{\pi}{2} \in \left[\pi, 3\frac{\pi}{2}\right] ~\forall \theta \in \left[\frac{\pi}{2}, \pi\right].$$

The last shows that $$\alpha = \theta + \frac{\pi}{2}$$ does not work.

You would be correct in saying that trigonometric identities lead to an ambiguity. If we apply the "angle-addition" formula for cotangent (so that we don't have to deal with undefined values), we have $$\cot ( \ \theta \ \pm \ \phi \ ) \ \ = \ \ \frac{\cot \theta · \cot \phi \ \mp \ 1}{ \cot \phi \ \pm \ \cot \theta} \ \ ;$$ since your result for the argument of $$\ z \$$ is $$\ \cot \alpha \ = \ -\frac{1}{\cot \theta} \ \ , \$$ we find that $$\ \cot \frac{\pi}{2} \ = \ 0 \ \$$ produces $$\cot \left( \ \theta \ \pm \ \frac{\pi}{2} \ \right) \ \ = \ \ \frac{\cot \theta · 0 \ \mp \ 1}{ 0 \ \pm \ \cot \theta} \ \ = \ \ -\frac{ 1}{ \cot \theta} \ \ .$$ So on these grounds alone, it seems that we should have $$\ \alpha \ = \ \theta \ \pm \ \frac{\pi}{2} \ \ .$$

However, the given expression $$\ z \ = \ ( \ 1 - \cos [2 \theta] \ ) - i·\sin [2 \theta] \ \$$ should be considered in terms of where the numbers described lie on the Argand plane. (We will see below that this is a circle of radius $$\ 2 \$$ centered on $$\ (1 \ , \ 0) \ \ , \$$ which is "traced-out" counter-clockwise starting at the origin.)

The way we'll discuss this is to examine transformations of $$\ u \ = \ \cos \theta \ + \ i·\sin \theta \ \ , \$$ with $$\ |u| \ = \ 1 \ \ , \$$ over $$\ 0 \ \le \ \theta \ \le \ \pi \ \ . \$$ [This is marked on the graph at left below along the red semi-circular arc, with "landmark" values shown, including $$\ v \ = \ cis \ \frac{\pi}{4} \ = \ \frac{\sqrt2}{2} + i·\frac{\sqrt2}{2} \$$ and $$\ w \ = \ cis \ \frac{3 \pi}{4} \ = \ -\frac{\sqrt2}{2} + i·\frac{\sqrt2}{2} \ \ . \ ] \$$ DeMoivre tells us that $$\ u^2 \ = \ \cos (2 \theta) \ + \ i·\sin (2 \theta) \ \ , \$$ the transformed numbers still having unit modulus and are now arranged counter-clockwise over the full unit circle [in orange]. Multiplying these numbers by $$\ -1 \$$ to produce $$\ -u^2 \$$ "reverses" their directions relative to the origin, or can be interpreted as having $$\ \pi \$$ added to their arguments, since $$\ -1·u^2 \ = \ e^{ \ i \pi}·e^{ \ i \ · \ 2 \theta} \ \ ; \$$ this has no effect on the "orientation" of the numbers on the circle [green in the graph at right below]. (Again, the "labeled" numbers are the values of $$\ u \$$ marked at the "positions" to which they have been transformed.)

The remaining transformation to obtain $$\ z \ = \ 1 - ( \ \cos [2 \theta] \ + i·\sin [2 \theta] \ ) \ = \ 1 - u^2 \ \$$ adds $$\ 1 \$$ to the real part of $$\ -u^2 \ \ , \$$ which has the effect of horizontally translating the circle "to the right" by $$\ 1 \$$ unit [seen in the graph below]. This final transformed circle (in fact, a "one-petal rose") can be represented, using $$\ \alpha \$$ in the manner marked, as $$\ r \ = \ 2 \cos \alpha \$$ over $$\ -\frac{\pi}{2} \ \le \ \alpha \ \le \ +\frac{\pi}{2} \ \ , \$$ with $$\ \alpha \$$ behaving as a "standard" polar angle. Using the argument of $$\ u \$$ instead, the circle is also represented by $$\ r \ = \ 2 \sin \theta \ = \ |z| \ \$$ over $$\ 0 \ \le \ \theta \ \le \ \pi \ \ ; \$$ so this circle can be interpreted as displaying the modulus of $$\ z \ \ , \$$ which you found correctly. The angle variables increase in the counter-clockwise direction on both circles, making this consistent with $$\ \alpha \ = \ \theta \ - \ \frac{\pi}{2} \ \ . \$$

Having said all of this, we still have a source of ambiguity that would permit $$\ \alpha \ = \ \theta \ + \ \frac{\pi}{2} \ \$$ to be correct. After all, we could also have read $$\ -u^2 \$$ as $$\ -1·u^2 \ = \ \large{e^{ \ -i \pi} } \normalsize{·e^{ \ i \ · \ 2 \theta} \ \ .} \$$ This would place the unit-modulus numbers in the same positions on the transformed circle, but would reduce the value of $$\ \theta \$$ relative to $$\ \alpha \$$ by $$\ \pi \$$ radians. The "correctness" of the given answer is then owed to the "right-handed"/counter-clockwise convention of describing angles.