Question

$\left|\frac{\cos\alpha+i\sin\alpha}{\sin\alpha-i\cos\alpha}\right|^{1000}+\left|\frac{\sin\alpha+i\cos\alpha}{\cos\alpha-i\sin\alpha}\right|^{2000}$ is equal to

• A. 1
• B. 5
• C. 4
• D. 16
• E. 2

Hint:

Math Tip: Whenever you see an expression involving $(\sin\theta \pm i\cos\theta)$ or $(\cos\theta \pm i\sin\theta)$, check if applying the modulus property simplifies the problem. In this case, since $\sin^2\theta + \cos^2\theta = 1$, all individual moduli simply evaluate to 1.

Answer 1

Answer: (E)

Concept:
Complex Numbers - Modulus Properties.
The modulus of a quotient is the quotient of the moduli: $\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|}$.
The modulus of a complex number $z = a+bi$ is $|z| = \sqrt{a^2 + b^2}$.
Step 1: Evaluate the modulus of the first fraction.
Let's find the modulus of the numerator and denominator separately. $$ \text{Numerator: } |\cos\alpha + i\sin\alpha| = \sqrt{\cos^2\alpha + \sin^2\alpha} = \sqrt{1} = 1 $$ $$ \text{Denominator: } |\sin\alpha - i\cos\alpha| = \sqrt{\sin^2\alpha + (-\cos\alpha)^2} = \sqrt{\sin^2\alpha + \cos^2\alpha} = \sqrt{1} = 1 $$ Therefore, the modulus of the first fraction is: $$ \left|\frac{\cos\alpha + i\sin\alpha}{\sin\alpha - i\cos\alpha}\right| = \frac{1}{1} = 1 $$
Step 2: Evaluate the modulus of the second fraction.
Repeat the same process for the second term. $$ \text{Numerator: } |\sin\alpha + i\cos\alpha| = \sqrt{\sin^2\alpha + \cos^2\alpha} = \sqrt{1} = 1 $$ $$ \text{Denominator: } |\cos\alpha - i\sin\alpha| = \sqrt{\cos^2\alpha + (-\sin\alpha)^2} = \sqrt{\cos^2\alpha + \sin^2\alpha} = \sqrt{1} = 1 $$ Therefore, the modulus of the second fraction is: $$ \left|\frac{\sin\alpha + i\cos\alpha}{\cos\alpha - i\sin\alpha}\right| = \frac{1}{1} = 1 $$
Step 3: Substitute and evaluate the final expression.
Now substitute the evaluated moduli back into the original expression with their exponents: $$ (1)^{1000} + (1)^{2000} $$ Since $1$ raised to any power is $1$: $$ 1 + 1 = 2 $$