Argument of complex numbers Questions in English

(D) Given that $|{z_1} + {z_2}| = |{z_1}| + |{z_2}|$.
This condition implies that the complex numbers ${z_1}$ and ${z_2}$ lie on the same ray originating from the origin in the complex plane.
Let ${z_1} = {r_1}(\cos{\theta_1} + i\sin{\theta_1})$ and ${z_2} = {r_2}(\cos{\theta_2} + i\sin{\theta_2})$.
Then $|{z_1} + {z_2}|^2 = (|{z_1}| + |{z_2}|)^2 = |{z_1}|^2 + |{z_2}|^2 + 2|{z_1}||{z_2}|$.
Also,$|{z_1} + {z_2}|^2 = (z_1 + z_2)(\overline{z_1} + \overline{z_2}) = |z_1|^2 + |z_2|^2 + z_1\overline{z_2} + \overline{z_1}z_2 = |z_1|^2 + |z_2|^2 + 2\text{Re}(z_1\overline{z_2})$.
Comparing these,$2\text{Re}(z_1\overline{z_2}) = 2|z_1||z_2|$,which means $\cos(\theta_1 - \theta_2) = 1$.
Thus,$\theta_1 - \theta_2 = 0$,which implies $\text{arg}({z_1}) - \text{arg}({z_2}) = 0$.

(D) Let $z = 5 - \sqrt{3}i$.
The argument of a complex number $z = x + iy$ is given by $\theta = \tan^{-1}\left(\frac{y}{x}\right)$.
Here,$x = 5$ and $y = -\sqrt{3}$.
Therefore,$\theta = \tan^{-1}\left(\frac{-\sqrt{3}}{5}\right) = \tan^{-1}\left(-\frac{\sqrt{3}}{5}\right)$.
Thus,the correct option is $D$.

(C) Given $|z| = 4$ and $\text{arg}(z) = \frac{5\pi}{6} = 150^{\circ}$.
Let $z = x + iy = r(\cos \theta + i \sin \theta)$,where $r = |z| = 4$ and $\theta = \frac{5\pi}{6}$.
Then $x = r \cos \theta = 4 \cos(150^{\circ}) = 4 \times (-\frac{\sqrt{3}}{2}) = -2\sqrt{3}$.
And $y = r \sin \theta = 4 \sin(150^{\circ}) = 4 \times (\frac{1}{2}) = 2$.
Therefore,$z = -2\sqrt{3} + 2i$.

(C) Given $z = \frac{1 - i\sqrt{3}}{1 + i\sqrt{3}}$.
Multiply the numerator and denominator by the conjugate of the denominator $(1 - i\sqrt{3})$:
$z = \frac{(1 - i\sqrt{3})(1 - i\sqrt{3})}{(1 + i\sqrt{3})(1 - i\sqrt{3})} = \frac{1 - 3 - 2i\sqrt{3}}{1 + 3} = \frac{-2 - 2i\sqrt{3}}{4} = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$.
Since the complex number $z = x + iy$ lies in the third quadrant $(x < 0, y < 0)$,the argument is given by $\pi + \tan^{-1}(\frac{y}{x})$.
$arg(z) = 180^\circ + \tan^{-1}\left(\frac{-\sqrt{3}/2}{-1/2}\right) = 180^\circ + \tan^{-1}(\sqrt{3}) = 180^\circ + 60^\circ = 240^\circ$.
Alternatively,using the property $arg(\frac{z_1}{z_2}) = arg(z_1) - arg(z_2)$:
$arg(z) = arg(1 - i\sqrt{3}) - arg(1 + i\sqrt{3}) = -60^\circ - 60^\circ = -120^\circ$.
Since $-120^\circ$ is equivalent to $360^\circ - 120^\circ = 240^\circ$,the correct answer is $240^\circ$.

(B) Let $z = r(\cos \theta + i \sin \theta)$.
Then,the conjugate of $z$ is $\overline{z} = r(\cos \theta - i \sin \theta)$.
Using the property $\cos(-\theta) = \cos \theta$ and $\sin(-\theta) = -\sin \theta$,we can write $\overline{z} = r(\cos(-\theta) + i \sin(-\theta))$.
Therefore,$arg(\overline{z}) = -\theta$.

(B) Given $z = \sin \alpha i(1 - \cos \alpha )$.
The amplitude (argument) of $z = x iy$ is given by $\theta = \tan^{-1}\left(\frac{y}{x}\right)$.
Here,$x = \sin \alpha$ and $y = 1 - \cos \alpha$.
Therefore,$\text{amp}(z) = \tan^{-1}\left(\frac{1 - \cos \alpha}{\sin \alpha}\right)$.
Using trigonometric identities $1 - \cos \alpha = 2\sin^2\left(\frac{\alpha}{2}\right)$ and $\sin \alpha = 2\sin\left(\frac{\alpha}{2}\right)\cos\left(\frac{\alpha}{2}\right)$:
$\text{amp}(z) = \tan^{-1}\left(\frac{2\sin^2\left(\frac{\alpha}{2}\right)}{2\sin\left(\frac{\alpha}{2}\right)\cos\left(\frac{\alpha}{2}\right)}\right) = \tan^{-1}\left(\tan\left(\frac{\alpha}{2}\right)\right) = \frac{\alpha}{2}$.

(A) Let $z = \frac{1 + i\sqrt{3}}{\sqrt{3} + 1}$.
Since the real part is $\frac{1}{\sqrt{3} + 1}$ and the imaginary part is $\frac{\sqrt{3}}{\sqrt{3} + 1}$,both are positive.
Thus,$z$ lies in the first quadrant.
The argument (amplitude) is given by $\theta = \tan^{-1}\left(\frac{\text{Im}(z)}{\text{Re}(z)}\right)$.
$\theta = \tan^{-1}\left(\frac{\sqrt{3}/(\sqrt{3} + 1)}{1/(\sqrt{3} + 1)}\right) = \tan^{-1}(\sqrt{3})$.
Therefore,$\theta = \frac{\pi}{3}$.

(C) Let $z = -1 + i\sqrt{3}$.
Here,the real part $x = -1$ and the imaginary part $y = \sqrt{3}$.
Since $x < 0$ and $y > 0$,the complex number lies in the second quadrant.
The argument $\theta$ is given by $\theta = 180^\circ - \tan^{-1}\left|\frac{y}{x}\right|$.
$\theta = 180^\circ - \tan^{-1}\left|\frac{\sqrt{3}}{-1}\right| = 180^\circ - \tan^{-1}(\sqrt{3}) = 180^\circ - 60^\circ = 120^\circ$.

(C) Let $z = \frac{3 + i}{2 - i} + \frac{3 - i}{2 + i}$.
First,simplify the expression by finding a common denominator:
$z = \frac{(3 + i)(2 + i) + (3 - i)(2 - i)}{(2 - i)(2 + i)}$
$z = \frac{(6 + 3i + 2i + i^2) + (6 - 3i - 2i + i^2)}{4 - i^2}$
Since $i^2 = -1$,we have:
$z = \frac{(6 + 5i - 1) + (6 - 5i - 1)}{4 - (-1)}$
$z = \frac{5 + 5i + 5 - 5i}{5} = \frac{10}{5} = 2$.
Now,find the argument of $z = 2$:
$arg(2) = 0$ (since $2$ is a positive real number).
Therefore,the correct option is $C$.

(A) We know that the argument of a product of complex numbers is given by $arg(z_1 \cdot z_2 \cdot \dots \cdot z_n) = arg(z_1) + arg(z_2) + \dots + arg(z_n) + 2k\pi$,where $k$ is an integer.
Therefore,the sum $\sum_{i=1}^{n} arg(z_i)$ and $arg(z)$ differ by an integer multiple of $2\pi$.

(B) Let $z = 0 + ib$,where $b > 0$.
Since $z$ is purely imaginary and its imaginary part is positive,it lies on the positive $y$-axis in the complex plane.
The argument of any complex number lying on the positive $y$-axis is $\frac{\pi}{2}$.

(D) Let $z = 0 + ib$,where $b < 0$.
Since $z$ is a purely imaginary number with a negative imaginary part,it lies on the negative $y$-axis.
The argument of any complex number lying on the negative $y$-axis is $-\frac{\pi}{2}$.
Therefore,$\arg(z) = -\frac{\pi}{2}$.

(A) Let $z = a + i0$,where $a < 0$.
Since $z$ is a purely real number with a negative real part,it lies on the negative $x$-axis in the complex plane.
The argument of any point on the negative $x$-axis is $\pi$.
Therefore,$\text{arg}(z) = \pi$.

(C) We know that $|z_1 - z_2|^2 = |z_1|^2 + |z_2|^2 - 2|z_1||z_2| \cos(\theta_1 - \theta_2)$,where $\theta_1 = \arg(z_1)$ and $\theta_2 = \arg(z_2)$.
Given that $\arg(z_1/z_2) = 0$,we have $\arg(z_1) - \arg(z_2) = 0$,which implies $\theta_1 - \theta_2 = 0$.
Substituting this into the equation:
$|z_1 - z_2|^2 = |z_1|^2 + |z_2|^2 - 2|z_1||z_2| \cos(0)$
$|z_1 - z_2|^2 = |z_1|^2 + |z_2|^2 - 2|z_1||z_2|$
$|z_1 - z_2|^2 = (|z_1| - |z_2|)^2$
Taking the square root on both sides,we get $|z_1 - z_2| = ||z_1| - |z_2||$.

(C) Let $z = x + iy$. Given $0 < \text{amp}(z) < \pi$,the complex number $z$ lies in the upper half-plane.
Since $\text{amp}(z) = \theta$,we have $\text{amp}(-z) = \text{amp}(-(x + iy)) = \text{amp}(-x - iy)$.
Since $z$ is in the first or second quadrant,$-z$ is in the third or fourth quadrant.
Specifically,if $\text{amp}(z) = \theta$,then $\text{amp}(-z) = \theta - \pi$ (if $\theta > 0$) or $\theta + \pi$ (if $\theta < 0$).
Given $0 < \text{amp}(z) < \pi$,we have $\text{amp}(-z) = \text{amp}(z) - \pi$.
Therefore,$\text{amp}(z) - \text{amp}(-z) = \text{amp}(z) - (\text{amp}(z) - \pi) = \pi$.

(D) Given $z = 1 - \cos \alpha + i \sin \alpha$.
Using trigonometric identities $1 - \cos \alpha = 2 \sin^2 \frac{\alpha}{2}$ and $\sin \alpha = 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}$.
$z = 2 \sin^2 \frac{\alpha}{2} + i (2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}) = 2 \sin \frac{\alpha}{2} (\sin \frac{\alpha}{2} + i \cos \frac{\alpha}{2})$.
Since $\sin \frac{\alpha}{2} + i \cos \frac{\alpha}{2} = \cos(\frac{\pi}{2} - \frac{\alpha}{2}) + i \sin(\frac{\pi}{2} - \frac{\alpha}{2})$.
Thus,$\text{amp } z = \frac{\pi}{2} - \frac{\alpha}{2}$.

(C) We know that $\text{amp}\left( \frac{z_1}{z_2} \right) = \text{amp}(z_1) - \text{amp}(z_2)$.
Also,$\text{amp}(\bar{z}) = -\text{amp}(z)$.
Therefore,$\text{amp}\left( \frac{z_1}{\bar{z}_2} \right) = \text{amp}(z_1) - \text{amp}(\bar{z}_2)$.
Since $\text{amp}(\bar{z}_2) = -\text{amp}(z_2)$,we get $\text{amp}(z_1) - (-\text{amp}(z_2)) = \text{amp}(z_1) + \text{amp}(z_2) = \text{amp}(z_1 z_2)$.
Comparing this with the given options,option $(C)$ is the correct expression.

(B) Let $z = \frac{13 - 5i}{4 - 9i}$.
To simplify,multiply the numerator and denominator by the conjugate of the denominator $(4 + 9i)$:
$z = \frac{(13 - 5i)(4 + 9i)}{(4 - 9i)(4 + 9i)}$
$z = \frac{52 + 117i - 20i - 45i^2}{16 + 81}$
$z = \frac{52 + 97i + 45}{97} = \frac{97 + 97i}{97} = 1 + i$
The argument of $z = 1 + i$ is $\theta = \tan^{-1}(\frac{1}{1}) = \frac{\pi}{4}$.

(A) Let $z = \frac{1 - i}{1 + i}$.
To simplify,multiply the numerator and denominator by the conjugate of the denominator $(1 - i)$:
$z = \frac{1 - i}{1 + i} \times \frac{1 - i}{1 - i} = \frac{(1 - i)^2}{1^2 - i^2} = \frac{1 - 2i + i^2}{1 - (-1)} = \frac{1 - 2i - 1}{2} = \frac{-2i}{2} = -i$.
The complex number is $z = 0 - i$.
Since the real part is $0$ and the imaginary part is $-1$,the point lies on the negative imaginary axis.
The amplitude of a complex number $z = x + iy$ where $x=0$ and $y < 0$ is $-\frac{\pi}{2}$.

(A) Let $z = \frac{1 + \sqrt{3}i}{\sqrt{3} + i}$.
Using the property of arguments,$\text{amp}\left(\frac{z_1}{z_2}\right) = \text{amp}(z_1) - \text{amp}(z_2)$.
$\text{amp}(1 + \sqrt{3}i) = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3}$.
$\text{amp}(\sqrt{3} + i) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$.
Therefore,$\text{amp}(z) = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$.

(D) The complex number $z = 0$ can be written as $0 + 0i$.
For any complex number $z = x + iy$,the amplitude (or argument) is defined as $\theta = \tan^{-1}(y/x)$ for $z \neq 0$.
Since $x = 0$ and $y = 0$,the ratio $y/x$ is $0/0$,which is indeterminate.
Therefore,the amplitude of $0$ is not defined.

(D) Let $z = \frac{1 + \sqrt{3}i}{\sqrt{3} - i}$.
Multiply the numerator and denominator by the conjugate of the denominator,$\sqrt{3} + i$:
$z = \frac{1 + \sqrt{3}i}{\sqrt{3} - i} \times \frac{\sqrt{3} + i}{\sqrt{3} + i}$
$z = \frac{\sqrt{3} + i + 3i + \sqrt{3}i^2}{3 - i^2}$
Since $i^2 = -1$,we have:
$z = \frac{\sqrt{3} + 4i - \sqrt{3}}{3 + 1} = \frac{4i}{4} = i$
Now,$z = 0 + 1i$. The amplitude (argument) of $z = i$ is $\theta = \tan^{-1}(\frac{1}{0}) = \pi / 2$.

(C) Given $z = \frac{-2}{1 + \sqrt{3}i}$.
Multiply the numerator and denominator by the conjugate $(1 - \sqrt{3}i)$:
$z = \frac{-2(1 - \sqrt{3}i)}{(1 + \sqrt{3}i)(1 - \sqrt{3}i)} = \frac{-2 + 2\sqrt{3}i}{1 + 3} = \frac{-2 + 2\sqrt{3}i}{4}$.
$z = -\frac{1}{2} + \frac{\sqrt{3}}{2}i$.
Since the real part is negative and the imaginary part is positive,$z$ lies in the second quadrant.
$arg(z) = \pi - \tan^{-1}\left|\frac{\sqrt{3}/2}{-1/2}\right| = \pi - \tan^{-1}(\sqrt{3}) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.

(C) Let $z = \sin \frac{\pi}{5} + i(1 - \cos \frac{\pi}{5})$.
Using trigonometric identities $\sin 2\theta = 2\sin \theta \cos \theta$ and $1 - \cos 2\theta = 2\sin^2 \theta$,we have:
$z = 2\sin \frac{\pi}{10} \cos \frac{\pi}{10} + i(2\sin^2 \frac{\pi}{10})$
$z = 2\sin \frac{\pi}{10} (\cos \frac{\pi}{10} + i\sin \frac{\pi}{10})$
Since $2\sin \frac{\pi}{10} > 0$,the argument (amplitude) $\theta$ is given by:
$\tan \theta = \frac{\sin(\pi/10)}{\cos(\pi/10)} = \tan \frac{\pi}{10}$
Therefore,$\theta = \frac{\pi}{10}$.

(D) Let $z = -1 - i\sqrt{3}$.
Here,the real part $a = -1$ and the imaginary part $b = -\sqrt{3}$.
First,we find the reference angle $\alpha = \tan^{-1}\left|\frac{b}{a}\right| = \tan^{-1}\left|\frac{-\sqrt{3}}{-1}\right| = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$.
Since both $a < 0$ and $b < 0$,the complex number $z$ lies in the $III$ quadrant.
The argument $\theta$ for a complex number in the $III$ quadrant is given by $\theta = -(\pi - \alpha)$.
Therefore,$\theta = -(\pi - \frac{\pi}{3}) = -(\frac{2\pi}{3}) = -\frac{2\pi}{3}$.

(B) Let $z = r(\cos \theta + i \sin \theta)$,where $\theta = \arg(z)$.
Let the other complex number be $w = R(\cos \phi + i \sin \phi)$,where $\phi = \arg(w)$.
Given that $\arg(z) + \arg(w) = \pi$,we have $\theta + \phi = \pi$,which implies $\phi = \pi - \theta$.
Thus,$w = R(\cos(\pi - \theta) + i \sin(\pi - \theta)) = R(-\cos \theta + i \sin \theta)$.
If we consider $z = x + iy$,then $\bar{z} = x - iy = r(\cos \theta - i \sin \theta)$.
Then $-\bar{z} = -x + iy = r(-\cos \theta + i \sin \theta)$.
Comparing this with $w$,we see that $w = -\bar{z}$ (assuming $R=r$).

(C) Given the complex number $z = -1 + i\sqrt{3}$.
We represent this in polar form as $z = r(\cos \theta + i\sin \theta) = re^{i\theta}$.
Comparing the real and imaginary parts:
$r\cos \theta = -1$ and $r\sin \theta = \sqrt{3}$.
Since the real part is negative and the imaginary part is positive,the complex number lies in the second quadrant.
$\tan \theta = \frac{r\sin \theta}{r\cos \theta} = \frac{\sqrt{3}}{-1} = -\sqrt{3}$.
In the second quadrant,$\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.
Thus,$\theta = \frac{2\pi}{3}$.

(B) Let $z = {e^{{e^{ - i\theta }}}} = {e^{\cos \theta - i\sin \theta }} = {e^{\cos \theta }}{e^{ - i\sin \theta }}$.
$z = {e^{\cos \theta }}[\cos (\sin \theta ) - i\sin (\sin \theta )]$.
$z = {e^{\cos \theta }}\cos (\sin \theta ) - i{e^{\cos \theta }}\sin (\sin \theta )$.
$amp(z) = {\tan ^{ - 1}}\left[ { - \frac{{{e^{\cos \theta }}\sin (\sin \theta )}}{{{e^{\cos \theta }}\cos (\sin \theta )}}} \right]$.
$amp(z) = {\tan ^{ - 1}}[\tan ( - \sin \theta )] = - \sin \theta $.

(D) Let $z = x + iy$. Then $z - a = (x - a) + iy$.
Given $arg(z - a) = \frac{\pi}{4}$.
This implies $\tan^{-1}\left(\frac{y}{x - a}\right) = \frac{\pi}{4}$ for $x > a$.
Therefore,$\frac{y}{x - a} = \tan\left(\frac{\pi}{4}\right) = 1$.
This simplifies to $y = x - a$,or $x - y - a = 0$,which represents a ray starting from $(a, 0)$ excluding the point $(a, 0)$ itself.
Since this is a part of a line,the locus is a straight line.

(D) Given expression: $\frac{4(\cos 75^o + i\sin 75^o)}{0.4(\cos 30^o + i\sin 30^o)}$
$= \frac{4}{0.4} \times \frac{\cos 75^o + i\sin 75^o}{\cos 30^o + i\sin 30^o}$
$= 10 [\cos(75^o - 30^o) + i\sin(75^o - 30^o)]$
$= 10(\cos 45^o + i\sin 45^o)$
$= 10(\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}})$
$= \frac{10}{\sqrt{2}}(1 + i)$

(B) Given $\sqrt{3} + i = (a + ib)(c + id)$.
Expanding the right side,we get $\sqrt{3} + i = (ac - bd) + i(ad + bc)$.
Comparing real and imaginary parts,we have $ac - bd = \sqrt{3}$ and $ad + bc = 1$.
We need to find the value of $\theta = \tan^{-1}\left(\frac{b}{a}\right) + \tan^{-1}\left(\frac{d}{c}\right)$.
Using the formula $\tan^{-1}(x) + \tan^{-1}(y) = n\pi + \tan^{-1}\left(\frac{x+y}{1-xy}\right)$,we have:
$\theta = n\pi + \tan^{-1}\left(\frac{\frac{b}{a} + \frac{d}{c}}{1 - \frac{b}{a} \cdot \frac{d}{c}}\right) = n\pi + \tan^{-1}\left(\frac{bc + ad}{ac - bd}\right)$.
Substituting the values,$\theta = n\pi + \tan^{-1}\left(\frac{1}{\sqrt{3}}\right)$.
Since $\tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$,the expression equals $n\pi + \frac{\pi}{6}$ for $n \in I$.

(A) Given that ${z_1}, {z_2}$ are conjugate,so ${z_2} = \overline{z_1}$.
Given that ${z_3}, {z_4}$ are conjugate,so ${z_4} = \overline{z_3}$.
We need to evaluate $arg\left( \frac{z_1}{z_4} \right) + arg\left( \frac{z_2}{z_3} \right)$.
Using the property $arg(a) + arg(b) = arg(ab)$,we get:
$arg\left( \frac{z_1}{z_4} \cdot \frac{z_2}{z_3} \right) = arg\left( \frac{z_1 z_2}{z_3 z_4} \right)$.
Since ${z_1} \overline{z_1} = |z_1|^2$,we have ${z_1} {z_2} = |z_1|^2$ and ${z_3} {z_4} = |z_3|^2$.
Substituting these,we get $arg\left( \frac{|z_1|^2}{|z_3|^2} \right) = arg\left( \left| \frac{z_1}{z_3} \right|^2 \right)$.
Since $\left| \frac{z_1}{z_3} \right|^2$ is a positive real number,its argument is $0$.

(C) Given that $\text{arg}(zw) = \pi$ $(i)$
$\overline{z} + i\overline{w} = 0$ $\Rightarrow \overline{z} = -i\overline{w}$ $\Rightarrow z = i w$ $\Rightarrow w = -iz$
Substitute $w = -iz$ into $(i)$:
$\text{arg}(z(-iz)) = \pi$
$\text{arg}(-iz^2) = \pi$
$\text{arg}(-i) + \text{arg}(z^2) = \pi$
$\text{arg}(-i) + 2\text{arg}(z) = \pi$
Since $\text{arg}(-i) = -\pi / 2$,we have:
$-\pi / 2 + 2\text{arg}(z) = \pi$
$2\text{arg}(z) = 3\pi / 2$
$\text{arg}(z) = 3\pi / 4$

(C) Given that $|z| = 1$ and $\text{arg}(z) = \theta$,we can write $z = e^{i\theta}$.
Since $|z| = 1$,we have $\bar{z} = \frac{1}{z}$.
Substituting this into the expression,we get:
$\frac{1+z}{1+\bar{z}} = \frac{1+z}{1+\frac{1}{z}} = \frac{1+z}{\frac{z+1}{z}} = z$.
Therefore,$\text{arg}\left( \frac{1+z}{1+\bar{z}} \right) = \text{arg}(z) = \theta$.

(C) The principal argument of a complex number $z$ lies in the interval $(-\pi, \pi]$.
Given that $\alpha$ and $\beta$ are the principal arguments of $z_1$ and $z_2$ respectively,the argument of the product $z_1 z_2$ is $\text{arg}(z_1 z_2) = \alpha + \beta$.
Since $-\pi < \alpha \le \pi$ and $-\pi < \beta \le \pi$,the sum $\alpha + \beta$ lies in the interval $(-2\pi, 2\pi]$.
We are given the condition $\alpha + \beta > \pi$.
To bring the argument back into the principal range $(-\pi, \pi]$,we subtract $2\pi$ from the sum.
Thus,the principal argument of $z_1 z_2$ is $\alpha + \beta - 2\pi$.

(C) Let $z = \sin \frac{6\pi}{5} + i(1 + \cos \frac{6\pi}{5})$.
Using the identities $\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$ and $1 + \cos \theta = 2 \cos^2 \frac{\theta}{2}$,we have:
$z = 2 \sin \frac{3\pi}{5} \cos \frac{3\pi}{5} + i(2 \cos^2 \frac{3\pi}{5})$
$z = 2 \cos \frac{3\pi}{5} (\sin \frac{3\pi}{5} + i \cos \frac{3\pi}{5})$
Since $\cos \frac{3\pi}{5}$ is negative,we write $z = -2 \cos \frac{3\pi}{5} (-\sin \frac{3\pi}{5} - i \cos \frac{3\pi}{5})$.
Using $-\sin \theta = \cos(\frac{\pi}{2} + \theta)$ and $-\cos \theta = \sin(\frac{\pi}{2} + \theta)$,we get:
$z = |z| (\cos(\frac{\pi}{2} + \frac{3\pi}{5}) + i \sin(\frac{\pi}{2} + \frac{3\pi}{5}))$
$z = |z| (\cos \frac{11\pi}{10} + i \sin \frac{11\pi}{10})$.
Alternatively,the argument $\theta$ satisfies $\tan \theta = \frac{1 + \cos(6\pi/5)}{\sin(6\pi/5)} = \frac{2 \cos^2(3\pi/5)}{2 \sin(3\pi/5) \cos(3\pi/5)} = \cot(3\pi/5) = \tan(\frac{\pi}{2} - \frac{3\pi}{5}) = \tan(-\frac{\pi}{10})$.
Since the real part is negative and the imaginary part is negative,the complex number lies in the $3^{rd}$ quadrant.
Thus,$\theta = \pi - \frac{\pi}{10} = \frac{9\pi}{10}$.

(B) Let $E = Arg\left( -i e^{i\frac{\pi}{9}} z^2 \right) + 2Arg\left( 2i e^{-i\frac{\pi}{18}} \bar{z} \right)$.
Using the property $Arg(z_1 z_2) = Arg(z_1) + Arg(z_2) \pmod{2\pi}$,we have:
$E = Arg(-i) + Arg(e^{i\frac{\pi}{9}}) + Arg(z^2) + 2[Arg(2i) + Arg(e^{-i\frac{\pi}{18}}) + Arg(\bar{z})]$.
Since $Arg(-i) = -\frac{\pi}{2}$,$Arg(e^{i\frac{\pi}{9}}) = \frac{\pi}{9}$,$Arg(z^2) = 2Arg(z)$,
$Arg(2i) = \frac{\pi}{2}$,$Arg(e^{-i\frac{\pi}{18}}) = -\frac{\pi}{18}$,and $Arg(\bar{z}) = -Arg(z)$:
$E = -\frac{\pi}{2} + \frac{\pi}{9} + 2Arg(z) + 2[\frac{\pi}{2} - \frac{\pi}{18} - Arg(z)]$.
$E = -\frac{\pi}{2} + \frac{\pi}{9} + 2Arg(z) + \pi - \frac{\pi}{9} - 2Arg(z)$.
$E = -\frac{\pi}{2} + \pi = \frac{\pi}{2}$.

(D) We know that $|z_1 + z_2|^2 = |z_1|^2 + |z_2|^2 + 2|z_1||z_2| \cos(\theta)$,where $\theta = |Arg(z_1) - Arg(z_2)|$.
Given $|z_1| = \sqrt{2}$,$|z_2| = \sqrt{3}$,and $|z_1 + z_2| = \sqrt{5 - 2\sqrt{3}}$.
Substituting these values into the formula:
$5 - 2\sqrt{3} = (\sqrt{2})^2 + (\sqrt{3})^2 + 2(\sqrt{2})(\sqrt{3}) \cos(\theta)$
$5 - 2\sqrt{3} = 2 + 3 + 2\sqrt{6} \cos(\theta)$
$5 - 2\sqrt{3} = 5 + 2\sqrt{6} \cos(\theta)$
$-2\sqrt{3} = 2\sqrt{6} \cos(\theta)$
$\cos(\theta) = \frac{-2\sqrt{3}}{2\sqrt{6}} = -\frac{\sqrt{3}}{\sqrt{6}} = -\frac{1}{\sqrt{2}}$
Since $\cos(\theta) = -\frac{1}{\sqrt{2}}$,we have $\theta = \frac{3\pi}{4}$.

(D) Let $z = x + iy$.
Given $|z + \bar{z}| = 4$,we have $|2x| = 4$,which implies $x = \pm 2$.
Given $|z - \bar{z}| = 2$,we have $|2iy| = 2$,which implies $|y| = 1$,so $y = \pm 1$.
The possible values for $z$ are $2+i, 2-i, -2+i, -2-i$.
The argument $\theta = Amp(z)$ satisfies $\tan \theta = \frac{y}{x} = \pm \frac{1}{2}$.
For $z = 2+i$,$\tan \theta = 1/2 \Rightarrow \theta \in (0, \pi/6)$.
For $z = 2-i$,$\tan \theta = -1/2 \Rightarrow \theta \in (-\pi/6, 0)$.
For $z = -2+i$,$\tan \theta = -1/2 \Rightarrow \theta \in (5\pi/6, \pi)$.
For $z = -2-i$,$\tan \theta = 1/2 \Rightarrow \theta \in (-\pi, -5\pi/6)$.
Comparing these with the options,the range $(\pi/6, \pi/4)$ is never attained by $Amp(z)$.

(A) Given that the modulus $r = |z| = 4$ and the argument $\theta = \text{arg}(z) = \frac{5\pi}{6}$.
The polar form of a complex number is given by $z = r(\cos \theta + i \sin \theta)$.
Substituting the given values:
$z = 4 \left( \cos \frac{5\pi}{6} + i \sin \frac{5\pi}{6} \right)$.
We know that $\cos \frac{5\pi}{6} = -\frac{\sqrt{3}}{2}$ and $\sin \frac{5\pi}{6} = \frac{1}{2}$.
Therefore,$z = 4 \left( -\frac{\sqrt{3}}{2} + i \frac{1}{2} \right)$.
$z = -2\sqrt{3} + 2i$.

(A) Let $z = r(\cos \theta + i \sin \theta)$,where $\theta = \arg(z)$.
Given $\arg(z) = \theta < 0$.
Then $-z = -r(\cos \theta + i \sin \theta) = r(\cos(\theta + \pi) + i \sin(\theta + \pi))$.
Thus,$\arg(-z) = \theta + \pi$ (if $\theta + \pi \le \pi$) or $\theta - \pi$ (if $\theta + \pi > \pi$).
Since $-\pi < \theta < 0$,we have $0 < \theta + \pi < \pi$.
Therefore,$\arg(-z) = \theta + \pi$.
Now,$\arg(-z) - \arg(z) = (\theta + \pi) - \theta = \pi$.

(C) Let $z = (1 + \cos 2\alpha) + i \sin 2\alpha$.
Using trigonometric identities,$1 + \cos 2\alpha = 2 \cos^2 \alpha$ and $\sin 2\alpha = 2 \sin \alpha \cos \alpha$.
So,$z = 2 \cos^2 \alpha + i(2 \sin \alpha \cos \alpha) = 2 \cos \alpha (\cos \alpha + i \sin \alpha)$.
Given $\frac{\pi}{2} < \alpha < \frac{3\pi}{2}$,$\cos \alpha$ is negative. Thus,the modulus $|z|$ must be positive.
$|z| = |2 \cos \alpha| = -2 \cos \alpha$.
To express $z$ in polar form $r(\cos \theta + i \sin \theta)$ where $r > 0$,we write:
$z = -2 \cos \alpha (-\cos \alpha - i \sin \alpha)$.
Since $-\cos \alpha = \cos(\alpha - \pi)$ and $-\sin \alpha = \sin(\alpha - \pi)$,
$z = -2 \cos \alpha (\cos(\alpha - \pi) + i \sin(\alpha - \pi))$.
Therefore,the modulus is $-2 \cos \alpha$ and the argument is $\alpha - \pi$.

(C) Two complex numbers $z_1 = a + ib$ and $z_2 = c + id$ are conjugates if $a = c$ and $b = -d$.
Given $z_1 = 5 + i(x^3y^2)$ and $z_2 = (x^3 + y^2) + 6i$.
Equating real and imaginary parts: $x^3 + y^2 = 5$ and $x^3y^2 = -6$.
Let $u = x^3$ and $v = y^2$. Then $u + v = 5$ and $uv = -6$.
The quadratic equation $t^2 - (u+v)t + uv = 0$ becomes $t^2 - 5t - 6 = 0$.
Solving for $t$: $(t - 6)(t + 1) = 0$,so $t = 6$ or $t = -1$.
Since $y^2 = v$ must be non-negative,we have $y^2 = 6$ and $x^3 = -1$,which implies $x = -1$.
Then $\tan \theta = \frac{y}{x} = \frac{\pm \sqrt{6}}{-1} = \mp \sqrt{6}$.
Therefore,$\tan^2 \theta = (\mp \sqrt{6})^2 = 6$.

(A) Consider the expression $\arg \left( \frac{z_1}{z_4} \right) + \arg \left( \frac{z_2}{z_3} \right)$.
Using the property $\arg \left( \frac{a}{b} \right) = \arg(a) - \arg(b)$,we get:
$= \arg(z_1) - \arg(z_4) + \arg(z_2) - \arg(z_3)$
Rearranging the terms,we have:
$= (\arg(z_1) + \arg(z_2)) - (\arg(z_3) + \arg(z_4))$
Given that $(z_1, z_2)$ and $(z_3, z_4)$ are pairs of complex conjugates,we have $z_2 = \bar{z}_1$ and $z_4 = \bar{z}_3$.
Substituting these into the expression:
$= (\arg(z_1) + \arg(\bar{z}_1)) - (\arg(z_3) + \arg(\bar{z}_3))$
Since $\arg(\bar{z}) = -\arg(z)$,we have:
$= (\arg(z_1) - \arg(z_1)) - (\arg(z_3) - \arg(z_3))$
$= 0 - 0 = 0$.

(B) Let $z = x + iy$,then $\bar{z} = x - iy$.
Given $z = 1 - \bar{z}$,we have $x + iy = 1 - (x - iy) = 1 - x + iy$.
Equating real parts,$x = 1 - x$,which gives $2x = 1$,so $x = \frac{1}{2}$.
Since $|z| = 1$,we have $x^2 + y^2 = 1$.
Substituting $x = \frac{1}{2}$,we get $\frac{1}{4} + y^2 = 1$,so $y^2 = \frac{3}{4}$,which gives $y = \pm \frac{\sqrt{3}}{2}$.
Thus,$z = \frac{1}{2} \pm i\frac{\sqrt{3}}{2}$.
Since $z$ has an imaginary part,Statement $1$ is false.
The principal argument $\theta$ is given by $\tan \theta = \frac{y}{x}$.
For $z = \frac{1}{2} + i\frac{\sqrt{3}}{2}$,$\theta = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$.
For $z = \frac{1}{2} - i\frac{\sqrt{3}}{2}$,$\theta = \tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3}$.
Since the question implies a specific value for the argument,and $\frac{\pi}{3}$ is a valid principal argument for one of the possible values of $z$,Statement $2$ is considered true in this context.
Therefore,Statement $1$ is false and Statement $2$ is true.

(A) The roots of the quadratic equation $x^2 + x + 1 = 0$ are the complex cube roots of unity,$\omega$ and $\omega^2$.
Given $z = 3 + 6iz_0^{81} - 3iz_0^{93}$.
Since $\omega^3 = 1$,we have $z_0^{81} = (z_0^3)^{27} = 1^{27} = 1$ and $z_0^{93} = (z_0^3)^{31} = 1^{31} = 1$.
Substituting these values into the expression for $z$:
$z = 3 + 6i(1) - 3i(1) = 3 + 3i$.
To find $\arg(z)$,we use the formula $\arg(x + iy) = \tan^{-1}(\frac{y}{x})$ for $x > 0$.
$\arg(z) = \tan^{-1}(\frac{3}{3}) = \tan^{-1}(1) = \frac{\pi}{4}$.

(C) Let $z = \frac{3 + i \sin \theta}{4 - i \cos \theta}$. For $z$ to be a real number,its imaginary part must be zero.
Multiply the numerator and denominator by the conjugate of the denominator $(4 + i \cos \theta)$:
$z = \frac{(3 + i \sin \theta)(4 + i \cos \theta)}{(4 - i \cos \theta)(4 + i \cos \theta)} = \frac{12 + 3i \cos \theta + 4i \sin \theta - \sin \theta \cos \theta}{16 + \cos^2 \theta}$.
The imaginary part is $\frac{3 \cos \theta + 4 \sin \theta}{16 + \cos^2 \theta} = 0$.
This implies $3 \cos \theta + 4 \sin \theta = 0$,so $\tan \theta = -\frac{3}{4}$.
Since $\tan \theta = -\frac{3}{4}$,$\sin \theta$ and $\cos \theta$ have opposite signs.
We want the argument of $w = \sin \theta + i \cos \theta$.
Since $\tan \theta = -\frac{3}{4}$,let $\alpha = \tan^{-1}(\frac{3}{4})$. Then $\theta$ is in the second or fourth quadrant.
If $\theta$ is in the second quadrant,$\sin \theta = \frac{3}{5}, \cos \theta = -\frac{4}{5}$. Then $w = \frac{3}{5} - i \frac{4}{5}$. The argument is $-\tan^{-1}(\frac{4}{3})$.
If $\theta$ is in the fourth quadrant,$\sin \theta = -\frac{3}{5}, \cos \theta = \frac{4}{5}$. Then $w = -\frac{3}{5} + i \frac{4}{5}$. The argument is $\pi - \tan^{-1}(\frac{4}{3})$.

(A) The given complex number is $z = \frac{-16}{1+i \sqrt{3}}$.
Multiply the numerator and denominator by the conjugate $1-i \sqrt{3}$:
$z = \frac{-16(1-i \sqrt{3})}{(1+i \sqrt{3})(1-i \sqrt{3})} = \frac{-16(1-i \sqrt{3})}{1^2 - (i \sqrt{3})^2} = \frac{-16(1-i \sqrt{3})}{1+3} = \frac{-16(1-i \sqrt{3})}{4} = -4 + i 4 \sqrt{3}$.
Let $z = r(\cos \theta + i \sin \theta)$,where $r = \sqrt{(-4)^2 + (4 \sqrt{3})^2} = \sqrt{16 + 48} = \sqrt{64} = 8$.
Then $\cos \theta = \frac{-4}{8} = -\frac{1}{2}$ and $\sin \theta = \frac{4 \sqrt{3}}{8} = \frac{\sqrt{3}}{2}$.
Since $\cos \theta < 0$ and $\sin \theta > 0$,$\theta$ lies in the second quadrant.
$\theta = \pi - \frac{\pi}{3} = \frac{2 \pi}{3}$.
Thus,the polar form is $8\left(\cos \frac{2 \pi}{3} + i \sin \frac{2 \pi}{3}\right)$.