Question

If $a = \cos \theta + i \sin \theta, b = \cos 2\theta - i \sin 2\theta$ and $c = \cos 3\theta + i \sin 3\theta$ such that if $\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = 0$, then

• A. $\theta = 2k\pi, k \in Z$
• B. $\theta = (2k+1)\pi, k \in Z$
• C. $\theta = (4k+1)\pi, k \in Z$
• D. None of these

Answer 1

Answer: (A)

(a) $\theta = 2k\pi, k \in Z$

Answer 2

The given complex numbers are: $a = \cos \theta + i \sin \theta = e^{i\theta}$ $b = \cos 2\theta - i \sin 2\theta = \cos (-2\theta) + i \sin (-2\theta) = e^{-i2\theta}$ $c = \cos 3\theta + i \sin 3\theta = e^{i3\theta}$

The determinant given is: $D = \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}$

This is a cyclic determinant. The value of this determinant is given by the formula: $D = 3abc - (a^3 + b^3 + c^3)$

We are given that $D = 0$. So, $3abc - (a^3 + b^3 + c^3) = 0$ This can be rewritten as $a^3 + b^3 + c^3 - 3abc = 0$.

We know the algebraic identity: $a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

So, the condition $D=0$ implies: $(a+b+c)(a^2+b^2+c^2-ab-bc-ca) = 0$

This means either $a+b+c=0$ or $a^2+b^2+c^2-ab-bc-ca=0$.

The second condition, $a^2+b^2+c^2-ab-bc-ca=0$, can be rewritten as: $\frac{1}{2}[(a-b)^2 + (b-c)^2 + (c-a)^2] = 0$ For this sum of squares to be zero, each term must be zero (since $a,b,c$ are complex numbers, this holds if they are real, or if we consider magnitude squared, but here it implies $a-b=0$, etc.). This implies $a-b=0$, $b-c=0$, and $c-a=0$, which means $a=b=c$.

Let's analyze the two possibilities:

Case 1: $a=b=c$ If $a=b$, then $e^{i\theta} = e^{-i2\theta}$. This implies $e^{i\theta} / e^{-i2\theta} = 1$, so $e^{i(\theta - (-2\theta))} = 1$. $e^{i3\theta} = 1$. This means $3\theta = 2k\pi$ for some integer $k \in Z$. So, $\theta = \frac{2k\pi}{3}$.

If $b=c$, then $e^{-i2\theta} = e^{i3\theta}$. This implies $e^{i3\theta} / e^{-i2\theta} = 1$, so $e^{i(3\theta - (-2\theta))} = 1$. $e^{i5\theta} = 1$. This means $5\theta = 2m\pi$ for some integer $m \in Z$. So, $\theta = \frac{2m\pi}{5}$.

For $a=b=c$ to hold, both conditions for $\theta$ must be satisfied: $\frac{2k\pi}{3} = \frac{2m\pi}{5}$ $\frac{k}{3} = \frac{m}{5}$ $5k = 3m$. Since 3 and 5 are coprime, $k$ must be a multiple of 3. Let $k = 3n$ for some integer $n \in Z$. Substituting $k=3n$ into $\theta = \frac{2k\pi}{3}$: $\theta = \frac{2(3n)\pi}{3} = 2n\pi$.

Let's check if $\theta = 2n\pi$ satisfies $a=b=c$: $a = e^{i(2n\pi)} = \cos(2n\pi) + i \sin(2n\pi) = 1 + i(0) = 1$. $b = e^{-i2(2n\pi)} = e^{-i4n\pi} = \cos(-4n\pi) + i \sin(-4n\pi) = 1 + i(0) = 1$. $c = e^{i3(2n\pi)} = e^{i6n\pi} = \cos(6n\pi) + i \sin(6n\pi) = 1 + i(0) = 1$. Indeed, $a=b=c=1$ when $\theta = 2n\pi$. This is a valid set of solutions. This matches option (a).

Case 2: $a+b+c=0$ $e^{i\theta} + e^{-i2\theta} + e^{i3\theta} = 0$ This implies $(\cos \theta + i \sin \theta) + (\cos 2\theta - i \sin 2\theta) + (\cos 3\theta + i \sin 3\theta) = 0$. Separating real and imaginary parts: $(\cos \theta + \cos 2\theta + \cos 3\theta) + i (\sin \theta - \sin 2\theta + \sin 3\theta) = 0$. For this to be true, both real and imaginary parts must be zero.

Real part: $\cos \theta + \cos 2\theta + \cos 3\theta = 0$ $(\cos \theta + \cos 3\theta) + \cos 2\theta = 0$ Using $\cos A + \cos B = 2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$: $2 \cos \left(\frac{\theta+3\theta}{2}\right) \cos \left(\frac{3\theta-\theta}{2}\right) + \cos 2\theta = 0$ $2 \cos 2\theta \cos \theta + \cos 2\theta = 0$ $\cos 2\theta (2 \cos \theta + 1) = 0$. This implies either $\cos 2\theta = 0$ or $2 \cos \theta + 1 = 0$.

Imaginary part: $\sin \theta - \sin 2\theta + \sin 3\theta = 0$ $(\sin \theta + \sin 3\theta) - \sin 2\theta = 0$ Using $\sin A + \sin B = 2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$: $2 \sin \left(\frac{\theta+3\theta}{2}\right) \cos \left(\frac{3\theta-\theta}{2}\right) - \sin 2\theta = 0$ $2 \sin 2\theta \cos \theta - \sin 2\theta = 0$ $\sin 2\theta (2 \cos \theta - 1) = 0$. This implies either $\sin 2\theta = 0$ or $2 \cos \theta - 1 = 0$.

Now we need to find $\theta$ that satisfies both conditions simultaneously:

1. $(\cos 2\theta = 0 \text{ or } 2 \cos \theta + 1 = 0)$
2. $(\sin 2\theta = 0 \text{ or } 2 \cos \theta - 1 = 0)$

Let's check the combinations: Subcase 2.1: $\cos 2\theta = 0$ If $\cos 2\theta = 0$, then $2\theta = (n+\frac{1}{2})\pi$, which implies $\sin 2\theta = \pm 1 \ne 0$. For the second equation $\sin 2\theta (2 \cos \theta - 1) = 0$ to hold, we must have $2 \cos \theta - 1 = 0$, so $\cos \theta = 1/2$. If $\cos \theta = 1/2$, then $\theta = 2k\pi \pm \frac{\pi}{3}$. Then $2\theta = 4k\pi \pm \frac{2\pi}{3}$. $\cos 2\theta = \cos(\pm \frac{2\pi}{3}) = -1/2$. This contradicts $\cos 2\theta = 0$. So, this subcase yields no solution.

Subcase 2.2: $2 \cos \theta + 1 = 0$ If $2 \cos \theta + 1 = 0$, then $\cos \theta = -1/2$. Then $2 \cos \theta - 1 = 2(-1/2) - 1 = -1 - 1 = -2 \ne 0$. For the second equation $\sin 2\theta (2 \cos \theta - 1) = 0$ to hold, we must have $\sin 2\theta = 0$. If $\sin 2\theta = 0$, then $2\theta = n\pi$, so $\theta = n\pi/2$. If $\theta = n\pi/2$, then $\cos \theta$ can be $0, 1, -1$. However, we need $\cos \theta = -1/2$. This is not possible for $\theta = n\pi/2$. So, this subcase also yields no solution.

Since both subcases for $a+b+c=0$ lead to contradictions, there are no solutions for $\theta$ when $a+b+c=0$.

Therefore, the only valid solutions come from $a=b=c$, which gives $\theta = 2k\pi$, where $k \in Z$.

The final answer is $\boxed{\text{(a)}}$.