Q.Convert $40^\circ\, 20'$ into radian measure.

The most effective way to handle powers of complex numbers, especially when they are in a fractional form, is to convert them into their polar form. This allows us to use De Moivre's Theorem, which simplifies the process of raising a complex number to a power and also makes division straightforward. We will then relate the final result to the complex cube roots of unity.

Here's how we approach this problem:

1. Convert the numerator to polar form: Let $z_1 = \sqrt{3}-i$. The real part is $x_1 = \sqrt{3}$ and the imaginary part is $y_1 = -1$. The modulus $r_1$ is given by $r_1 = \sqrt{x_1^2 + y_1^2}$.

The argument $\theta_1$ is found using $\tan\theta_1 = \frac{y_1}{x_1} = \frac{-1}{\sqrt{3}}$. Since $x_1 > 0$ and $y_1 < 0$, $z_1$ lies in the fourth quadrant.
The principal argument is $\theta_1 = -\frac{\pi}{6}$.
So, $z_1 = 2\left(\cos\left(-\frac{\pi}{6}\right) + i\sin\left(-\frac{\pi}{6}\right)\right)$.

2. Convert the denominator to polar form:

Let $z_2 = -\sqrt{2}+\sqrt{2}i$.

The real part is $x_2 = -\sqrt{2}$ and the imaginary part is $y_2 = \sqrt{2}$.

The modulus $r_2$ is given by $r_2 = \sqrt{x_2^2 + y_2^2}$.

The argument $\theta_2$ is found using $\tan\theta_2 = \frac{y_2}{x_2} = \frac{\sqrt{2}}{-\sqrt{2}} = -1$. Since $x_2 < 0$ and $y_2 > 0$, $z_2$ lies in the second quadrant.
The principal argument is $\theta_2 = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.
So, $z_2 = 2\left(\cos\left(\frac{3\pi}{4}\right) + i\sin\left(\frac{3\pi}{4}\right)\right)$.

3. Divide the complex numbers in polar form:

Let $Z = \frac{z_1}{z_2}$. When dividing complex numbers in polar form, we divide their moduli and subtract their arguments.

Calculate the argument:

So, $Z = \cos\left(-\frac{11\pi}{12}\right) + i\sin\left(-\frac{11\pi}{12}\right)$.

4. Apply De Moivre's Theorem for the power:

We need to calculate $Z^{20}$.

> [!FORMULA]

> De Moivre's Theorem states that for any complex number $z = r(\cos\theta + i\sin\theta)$ and any integer $n$,

> $z^n = r^n(\cos(n\theta) + i\sin(n\theta))$.

In our case, $r=1$ and $n=20$.

Simplify the angle:

To find the principal argument, we can add multiples of $2\pi$:

Since $\cos(\theta + 2k\pi) = \cos\theta$ and $\sin(\theta + 2k\pi) = \sin\theta$ for any integer $k$, we can write:

Using the identities $\cos(-x) = \cos x$ and $\sin(-x) = -\sin x$:

Substitute the values of $\cos(\pi/3)$ and $\sin(\pi/3)$:

5. Relate the result to complex cube roots of unity:

   The complex cube roots of unity are $1, \omega, \omega^2$.

   We know that $\omega = e^{i2\pi/3} = \cos\left(\frac{2\pi}{3}\right) + i\sin\left(\frac{2\pi}{3}\right) = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$.

   And $\omega^2 = e^{i4\pi/3} = \cos\left(\frac{4\pi}{3}\right) + i\sin\left(\frac{4\pi}{3}\right) = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$.

   Our result is $\frac{1}{2} - i\frac{\sqrt{3}}{2}$.

   Let's check the options:

   • (A) $\omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$
   • (B) $\omega - \omega^2 = \left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) - \left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) = i\sqrt{3}$
   • (C) $\omega^2 = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$
   • (D) $-\omega = -\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = \frac{1}{2} - i\frac{\sqrt{3}}{2}$

   Our calculated value matches $-\omega$.

The key idea is that when we need the value of a trigonometric function at an angle close to one we know exactly, we can use the linear approximation (or differentials).

Here, $59^\circ$ is just $1^\circ$ less than $60^\circ$, and we know $\sec 60^\circ = 2$ exactly.

The derivative of $\sec x$ tells us how fast $\sec x$ changes near $60^\circ$, so we can estimate the small drop from $2$ to $\sec 59^\circ$.

1. Convert the angle difference to radians Since calculus formulas require radians, we convert $1^\circ$ to radians using the given approximation:

So the change in angle is $\Delta x = -0.0174$ (negative because $59^\circ$ is smaller than $60^\circ$).

2. Recall the derivative of $\sec x$

This tells us the instantaneous rate of change of $\sec x$ at any $x$.

3. Evaluate the derivative at $x = 60^\circ$ (in radians) First, $60^\circ = \frac{\pi}{3}$ radians. We know:

Hence,

4. Apply the linear approximation formula For a small change $\Delta x$,

Here $\Delta x = -0.0174$, so

5. Compute the correction

First, $3.464 \times 1.74$:

• $3.464 \times 1.7 = 5.8888$
• $3.464 \times 0.04 = 0.13856$
• Sum = $6.02736$ So $3.464 \times 0.0174 \approx 0.06027$. Thus the correction is $-0.06027$.

6. Final approximation

Rounded to four decimal places, this is $1.9397$.