Question
Let $z = 1 + \cos 2\theta + i \sin 2\theta$, where $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$.
(a) Show that
(i) $\arg z = \theta$;
(ii) $|z| = 2 \cos \theta$.
(b) Hence or otherwise, find the value of $\theta$ such that $\arg(z^2) = |z|^3$.
▶️Answer/Explanation
Detailed Solution
(a)(i) Show that \(\arg z = \theta\)
To find the argument of \( z \), we need to express \( z \) in the form \( x + iy \), where \( x \) is the real part and \( y \) is the imaginary part, and then compute \( \arg z = \tan^{-1} \left( \frac{y}{x} \right) \), adjusting for the quadrant. Let’s start by simplifying the expression for \( z \):
\[
z = 1 + \cos 2\theta + i \sin 2\theta
\]
Real part: \( x = 1 + \cos 2\theta \)
Imaginary part: \( y = \sin 2\theta \)
Now, compute the argument:
\[
\arg z = \tan^{-1} \left( \frac{y}{x} \right) = \tan^{-1} \left( \frac{\sin 2\theta}{1 + \cos 2\theta} \right)
\]
This looks promising, but we need to show it equals \( \theta \). Let’s use trigonometric identities to simplify the expression. Recall the double-angle identities:
\( \sin 2\theta = 2 \sin \theta \cos \theta \)
\( \cos 2\theta = \cos^2 \theta – \sin^2 \theta = 2 \cos^2 \theta – 1 \)
Substitute these into the real and imaginary parts:
\[
x = 1 + \cos 2\theta = 1 + (2 \cos^2 \theta – 1) = 2 \cos^2 \theta
\]
\[
y = \sin 2\theta = 2 \sin \theta \cos \theta
\]
So:
\[
\frac{y}{x} = \frac{2 \sin \theta \cos \theta}{2 \cos^2 \theta} = \frac{\sin \theta}{\cos \theta} = \tan \theta
\]
Thus:
\[
\arg z = \tan^{-1} \left( \frac{2 \sin \theta \cos \theta}{2 \cos^2 \theta} \right) = \tan^{-1} (\tan \theta)
\]
Since \( -\frac{\pi}{2} < \theta < \frac{\pi}{2} \), and \( \tan^{-1} (\tan \theta) = \theta \) in this interval (where the tangent function is one-to-one), we have:
\[
\arg z = \theta
\]
We should verify the quadrant: \( x = 2 \cos^2 \theta > 0 \) (since \( \cos^2 \theta > 0 \)), and \( y = 2 \sin \theta \cos \theta \) has the sign of \( \sin \theta \) (since \( \cos \theta > 0 \) for \( -\frac{\pi}{2} < \theta < \frac{\pi}{2} \)). When \( \theta > 0 \), \( y > 0 \), so \( z \) is in the first quadrant; when \( \theta < 0 \), \( y < 0 \), fourth quadrant—consistent with \( \arg z = \theta \). This holds, so part (i) is shown.
(a)(ii) Show that \(|z| = 2 \cos \theta\)
The modulus of a complex number \( z = x + iy \) is:
\[
|z| = \sqrt{x^2 + y^2}
\]
Using our expressions:
\[
x = 1 + \cos 2\theta, \quad y = \sin 2\theta
\]
\[
|z| = \sqrt{(1 + \cos 2\theta)^2 + (\sin 2\theta)^2}
\]
Expand the expression:
\[
(1 + \cos 2\theta)^2 = 1 + 2 \cos 2\theta + \cos^2 2\theta
\]
\[
(\sin 2\theta)^2 = \sin^2 2\theta
\]
\[
|z|^2 = (1 + \cos 2\theta)^2 + (\sin 2\theta)^2 = 1 + 2 \cos 2\theta + \cos^2 2\theta + \sin^2 2\theta
\]
Since \( \cos^2 2\theta + \sin^2 2\theta = 1 \):
\[
|z|^2 = 1 + 2 \cos 2\theta + 1 = 2 + 2 \cos 2\theta = 2 (1 + \cos 2\theta)
\]
Now, use \( 1 + \cos 2\theta = 2 \cos^2 \theta \):
\[
|z|^2 = 2 \cdot 2 \cos^2 \theta = 4 \cos^2 \theta
\]
\[
|z| = \sqrt{4 \cos^2 \theta} = 2 \sqrt{\cos^2 \theta}
\]
Since \( -\frac{\pi}{2} < \theta < \frac{\pi}{2} \), \( \cos \theta > 0 \), so:
\[
|z| = 2 \cos \theta
\]
This matches perfectly, and \( |z| \geq 0 \), which is satisfied since \( \cos \theta > 0 \) in the interval. Part (ii) is shown.
(b) Find the value of \(\theta\) such that \(\arg(z^2) = |z|^3\)
Using the results from part (a):
\( \arg z = \theta \)
\( |z| = 2 \cos \theta \)
First, compute \( \arg(z^2) \):
For a complex number \( z \), \( \arg(z^n) = n \arg z \)
\[
\arg(z^2) = 2 \arg z = 2 \theta
\]
Next, compute \( |z|^3 \):
\[
|z|^3 = (|z|)^3 = (2 \cos \theta)^3 = 8 \cos^3 \theta
\]
The equation becomes:
\[
\arg(z^2) = |z|^3
\]
\[
2 \theta = 8 \cos^3 \theta
\]
\[
\theta = 4 \cos^3 \theta
\]
Let’s solve this. Define \( f(\theta) = \theta – 4 \cos^3 \theta \), and find where \( f(\theta) = 0 \) in \( -\frac{\pi}{2} < \theta < \frac{\pi}{2} \).
– Test \( \theta = 0 \):
\[
f(0) = 0 – 4 \cos^3 0 = 0 – 4 \cdot 1 = -4 < 0
\]
Test \( \theta = \frac{\pi}{4} \):
\[
\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}, \quad \cos^3 \frac{\pi}{4} = \left(\frac{\sqrt{2}}{2}\right)^3 = \frac{\sqrt{2}}{2\sqrt{2}} = \frac{1}{2\sqrt{2}}
\]
\[
4 \cos^3 \frac{\pi}{4} = 4 \cdot \frac{1}{2\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \approx 1.414
\]
\[
f\left(\frac{\pi}{4}\right) = \frac{\pi}{4} – 1.414 \approx 0.785 – 1.414 = -0.629 < 0
\]
Test \( \theta = \frac{\pi}{2} \) (boundary, not included):
\[
\cos \frac{\pi}{2} = 0, \quad f\left(\frac{\pi}{2}\right) = \frac{\pi}{2} – 0 \approx 1.571 > 0
\]
– Test \( \theta = -\frac{\pi}{4} \):
\[
\cos \left(-\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}, \quad 4 \cos^3 \left(-\frac{\pi}{4}\right) = \sqrt{2}
\]
\[
f\left(-\frac{\pi}{4}\right) = -\frac{\pi}{4} – \sqrt{2} \approx -0.785 – 1.414 = -2.199 < 0
\]
Try positive values closer to zero:
\( \theta = 0.5 \):
\[
\cos 0.5 \approx 0.8776, \quad \cos^3 0.5 \approx 0.676, \quad 4 \cos^3 0.5 \approx 2.704
\]
\[
f(0.5) = 0.5 – 2.704 = -2.204 < 0
\]
\( \theta = 1 \):
\[
\cos 1 \approx 0.5403, \quad \cos^3 1 \approx 0.158, \quad 4 \cos^3 1 \approx 0.632
\]
\[
f(1) = 1 – 0.632 = 0.368 > 0
\]
A root exists between 0.5 and 1. Refine with \( \theta = 0.8 \):
\[
\cos 0.8 \approx 0.6967, \quad \cos^3 0.8 \approx 0.338, \quad 4 \cos^3 0.8 \approx 1.352
\]
\[
f(0.8) = 0.8 – 1.352 = -0.552 < 0
\]
\( \theta = 0.9 \):
\[
\cos 0.9 \approx 0.6216, \quad 4 \cos^3 0.9 \approx 0.960
\]
\[
f(0.9) = 0.9 – 0.960 = -0.060 < 0
\]
\( \theta = 0.95 \):
\[
\cos 0.95 \approx 0.5817, \quad 4 \cos^3 0.95 \approx 0.788
\]
\[
f(0.95) = 0.95 – 0.788 = 0.162 > 0
\]
The root is between 0.9 and 0.95. Since \( \cos^3 \theta \) decreases, and the left side is linear, there’s one positive solution. Numerically, \( \theta \approx 0.93 \) satisfies it closely.
………………….Markscheme……………………………….
Solution: –
(i) $arg~z = arctan(\frac{sin~2\theta}{1+cos~2\theta}) (tan(arg~z) = \frac{sin~2\theta}{1+cos~2\theta})$
uses $2sin\theta cos\theta$ in the numerator and any double angle identity for $cos~2\theta$ in the denominator
$arg~z = arctan(\frac{2sin\theta cos\theta}{2cos^{2}\theta}) (tan(arg~z) = \frac{2sin\theta cos\theta}{2cos^{2}\theta})$
$\Rightarrow arg~z = arctan(tan~\theta) (-\frac{\pi}{2}<\theta<\frac{\pi}{2})$
$= \theta$
(ii) attempts to express $|z|$ in the form $\sqrt{(Re~z)^{2}+(Im~z)^{2}}$
$|z| = \sqrt{(1+cos~2\theta)^{2}+sin^{2}2\theta}$
attempts to expand $(1+cos~2\theta)^{2}$ and then uses $cos^{2}2\theta+sin^{2}2\theta=1$ an attempt to simplify
$|z| = \sqrt{2+2cos~2\theta}$
$|z| = \sqrt{4cos^{2}\theta} (=2|cos~\theta|)$
$= 2cos~\theta (-\frac{\pi}{2}<\theta<\frac{\pi}{2})$
METHOD 2 (i) and (ii)
$z = 1 + 2\cos^2\theta – 1 + 2\sin\theta\cos\theta i$
$z = 2\cos^2\theta + 2\sin\theta\cos\theta i$
attempt to form $z = r cis\theta$
$z = 2\cos\theta(\cos\theta + i\sin\theta)$
$\therefore |z| = 2\cos\theta$ and $\arg z = \theta$.
(b) $2\theta = (2\cos\theta)^3$
attempts to solve for $\theta$
$\theta = 0.913236…$
$\theta = 0.913$