Key Concepts and Formulas

• Polar Form of a Complex Number: A complex number $z = x + iy$ can be represented in polar form as $z = r(\cos \theta + i \sin \theta)$, where $r = |z| = \sqrt{x^2 + y^2}$ is the modulus and $\theta = \arg(z)$ is the argument.
• Division of Complex Numbers in Polar Form: If $z_1 = r_1(\cos \theta_1 + i \sin \theta_1)$ and $z_2 = r_2(\cos \theta_2 + i \sin \theta_2)$, then $\frac{z_1}{z_2} = \frac{r_1}{r_2}(\cos(\theta_1 - \theta_2) + i \sin(\theta_1 - \theta_2))$.
• Argument of a Complex Number: The argument $\theta$ of a complex number $z = x + iy$ satisfies $\tan \theta = \frac{y}{x}$. The quadrant of $z$ determines the specific value of $\theta$.

Step-by-Step Solution

Step 1: Convert the Numerator to Polar Form

We are given the numerator $N = i - 1 = -1 + i$. We need to find its modulus and argument.

• Calculate the modulus ($r_N$): The modulus is the distance from the origin to the point $(-1, 1)$ in the complex plane. $r_N = |-1 + i| = \sqrt{(-1)^2 + (1)^2} = \sqrt{1 + 1} = \sqrt{2}$
• Calculate the argument ($\theta_N$): The complex number $-1 + i$ lies in the second quadrant. We first find the reference angle $\alpha$ such that $\tan \alpha = \left|\frac{1}{-1}\right| = 1$, so $\alpha = \frac{\pi}{4}$. Since the number is in the second quadrant, the argument is given by: $\theta_N = \pi - \alpha = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$
• Write the numerator in polar form: $N = \sqrt{2}\left(\cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4}\right)$

Step 2: Identify the Denominator's Polar Form

The denominator is given as $D = \cos \frac{\pi}{3} + i \sin \frac{\pi}{3}$.

• Recognize the polar form: This is already in the standard polar form $r(\cos \theta + i \sin \theta)$.
• Identify modulus and argument: By comparison, the modulus is $r_D = 1$ and the argument is $\theta_D = \frac{\pi}{3}$.

Step 3: Perform the Division using Polar Form

We have $z = \frac{N}{D}$, so we use the division rule for complex numbers in polar form: $z = \frac{r_N}{r_D} \left(\cos(\theta_N - \theta_D) + i \sin(\theta_N - \theta_D)\right)$

• Calculate the modulus of $z$ ($r$): $r = \frac{r_N}{r_D} = \frac{\sqrt{2}}{1} = \sqrt{2}$
• Calculate the argument of $z$ ($\theta$): $\theta = \theta_N - \theta_D = \frac{3\pi}{4} - \frac{\pi}{3} = \frac{9\pi}{12} - \frac{4\pi}{12} = \frac{5\pi}{12}$
• Write $z$ in polar form: $z = \sqrt{2}\left(\cos \frac{5\pi}{12} + i \sin \frac{5\pi}{12}\right)$

Step 4: Compare with Given Options

Our calculated value is $z = \sqrt{2}\left(\cos \frac{5\pi}{12} + i \sin \frac{5\pi}{12}\right)$. Comparing with the options: (A) $\cos \frac{\pi}{12}-i \sin \frac{\pi}{12}$ (B) $\sqrt{2}\left(\cos \frac{\pi}{12}+i \sin \frac{\pi}{12}\right)$ (C) $\sqrt{2} i\left(\cos \frac{5 \pi}{12}-i \sin \frac{5 \pi}{12}\right)$ (D) $\sqrt{2}\left(\cos \frac{5 \pi}{12}+i \sin \frac{5 \pi}{12}\right)$

Our result matches option (D). The given correct answer (A) is incorrect. The following steps will show how to obtain option A if numerator was $-1 - i$

Let's assume the numerator was actually $-1 - i$. The modulus is still $\sqrt{2}$. The argument is now $\pi + \frac{\pi}{4} = \frac{5\pi}{4}$. So $z = \frac{\sqrt{2}(\cos(5\pi/4) + i \sin(5\pi/4))}{\cos(\pi/3) + i \sin(\pi/3)} = \sqrt{2}(\cos(5\pi/4 - \pi/3) + i \sin(5\pi/4 - \pi/3)) = \sqrt{2}(\cos(11\pi/12) + i \sin(11\pi/12))$ This still doesn't correspond to option A.

Let's assume that the complex number is $z = \frac{1-i}{\cos(\pi/3) + i \sin(\pi/3)}$. Then numerator is $\sqrt{2}(\cos(-\pi/4) + i \sin(-\pi/4))$. $z = \sqrt{2}(\cos(-\pi/4 - \pi/3) + i \sin(-\pi/4 - \pi/3)) = \sqrt{2}(\cos(-7\pi/12) + i \sin(-7\pi/12))$. Still doesn't correspond to option A.

Let's work backwards from option A. If $z = \cos(\pi/12) - i \sin(\pi/12) = \cos(-\pi/12) + i \sin(-\pi/12)$. Then, $\frac{i-1}{z} = \cos(\pi/3) + i \sin(\pi/3)$. $i-1 = \sqrt{2}(\cos(3\pi/4) + i \sin(3\pi/4))$. So $\frac{\sqrt{2}(\cos(3\pi/4) + i \sin(3\pi/4))}{\cos(-\pi/12) + i \sin(-\pi/12)} = \cos(\pi/3) + i \sin(\pi/3)$. $\sqrt{2}(\cos(3\pi/4 + \pi/12) + i \sin(3\pi/4 + \pi/12)) = \cos(\pi/3) + i \sin(\pi/3)$. $\sqrt{2}(\cos(10\pi/12) + i \sin(10\pi/12)) = \cos(\pi/3) + i \sin(\pi/3)$. $\sqrt{2}(\cos(5\pi/6) + i \sin(5\pi/6)) = \cos(\pi/3) + i \sin(\pi/3)$. This is false.

Common Mistakes & Tips

• Quadrant Awareness: Always be mindful of the quadrant when determining the argument of a complex number. The arctangent function only provides the reference angle.
• Polar Form Standard: Ensure that you have correctly identified the modulus and argument from the polar form.
• Angle Arithmetic: Pay close attention when adding or subtracting angles, especially when finding a common denominator.

Summary

We converted the numerator and denominator of the given complex number into polar form and then applied the division rule for complex numbers in polar form. This led us to the result $z = \sqrt{2}\left(\cos \frac{5\pi}{12} + i \sin \frac{5\pi}{12}\right)$, which corresponds to option (D). However, the problem statement indicates that the correct answer is (A), suggesting a possible error in the problem statement or the provided answer key.

The final answer is \boxed{\sqrt{2}\left(\cos \frac{5 \pi}{12}+i \sin \frac{5 \pi}{12}\right)}, which corresponds to option (D).