Question

If \( \sqrt{-3 - 4i} = re^{i\theta} \), then \( r^2 \tan \theta = \)

• A. \( -5 \)
• B. \( 5 \)
• C. \( 10 \)
• D. \( -10 \)

Hint:

When dealing with square roots of complex numbers in polar form, squaring both sides helps to relate the magnitude and argument. Remember the double angle formulas for trigonometric functions when relating \( \theta \) and \( 2\theta \). Also, pay attention to the quadrant of the angle based on the signs of the real and imaginary parts.

Answer 1

The Correct Option is D

We are given \( \sqrt{-3 - 4i} = re^{i\theta} \).
Squaring both sides, we get: $$-3 - 4i = (re^{i\theta})^2 = r^2 (e^{i\theta})^2 = r^2 e^{2i\theta}$$ Using Euler's formula, \( e^{2i\theta} = \cos(2\theta) + i \sin(2\theta) \).
So, \( -3 - 4i = r^2 (\cos(2\theta) + i \sin(2\theta)) = r^2 \cos(2\theta) + i r^2 \sin(2\theta) \).
Equating the real and imaginary parts, we have: $$r^2 \cos(2\theta) = -3 \quad \cdots (1)$$ $$r^2 \sin(2\theta) = -4 \quad \cdots (2)$$ We need to find \( r^2 \tan \theta \).
We know that \( \tan \theta = \frac{\sin \theta}{\cos \theta} \).
We also know the double angle formulas for sine and cosine: $$ \sin(2\theta) = 2 \sin \theta \cos \theta $$ $$ \cos(2\theta) = \cos^2 \theta - \sin^2 \theta $$ From equations (1) and (2), we can find \( r^2 \): $$(r^2 \cos(2\theta))^2 + (r^2 \sin(2\theta))^2 = (-3)^2 + (-4)^2$$ $$r^4 (\cos^2(2\theta) + \sin^2(2\theta)) = 9 + 16$$ $$r^4 (1) = 25$$ $$r^4 = 25$$ Since \( r \) is the magnitude, \( r^2 = 5 \).
Now we have: $$5 \cos(2\theta) = -3 \implies \cos(2\theta) = -\frac{3}{5}$$ $$5 \sin(2\theta) = -4 \implies \sin(2\theta) = -\frac{4}{5}$$ We want to find \( r^2 \tan \theta = 5 \tan \theta \).
We know that \( \tan(2\theta) = \frac{\sin(2\theta)}{\cos(2\theta)} = \frac{-4/5}{-3/5} = \frac{4}{3} \).
Also, \( \tan(2\theta) = \frac{2 \tan \theta}{1 - \tan^2 \theta} \).
Let \( t = \tan \theta \).
$$\frac{4}{3} = \frac{2t}{1 - t^2}$$ $$4(1 - t^2) = 3(2t)$$ $$4 - 4t^2 = 6t$$ $$4t^2 + 6t - 4 = 0$$ $$2t^2 + 3t - 2 = 0$$ We can solve this quadratic equation for \( t \): $$t = \frac{-3 \pm \sqrt{3^2 - 4(2)(-2)}}{2(2)} = \frac{-3 \pm \sqrt{9 + 16}}{4} = \frac{-3 \pm \sqrt{25}}{4} = \frac{-3 \pm 5}{4}$$ So, \( t_1 = \frac{-3 + 5}{4} = \frac{2}{4} = \frac{1}{2} \) and \( t_2 = \frac{-3 - 5}{4} = \frac{-8}{4} = -2 \).
Since \( \cos(2\theta)<0 \) and \( \sin(2\theta)<0 \), \( 2\theta \) lies in the third quadrant.
In the third quadrant, \( \pi<2\theta<\frac{3\pi}{2} \), which implies \( \frac{\pi}{2}<\theta<\frac{3\pi}{4} \).
In this quadrant, \( \tan \theta \) is negative.
Therefore, \( \tan \theta = -2 \).
Finally, \( r^2 \tan \theta = 5 \times (-2) = -10 \).