Argument of complex numbers Questions in English

(N/A) Given $z = -1 - i \sqrt{3}$.
Let $z = r(\cos \theta + i \sin \theta)$,where $r$ is the modulus and $\theta$ is the argument.
Here,$r \cos \theta = -1$ and $r \sin \theta = -\sqrt{3}$.
Squaring and adding both equations:
$(r \cos \theta)^{2} + (r \sin \theta)^{2} = (-1)^{2} + (-\sqrt{3})^{2}$
$r^{2}(\cos^{2} \theta + \sin^{2} \theta) = 1 + 3$
$r^{2} = 4 \Rightarrow r = 2$ (since $r > 0$).
Thus,the modulus is $2$.
Now,$\cos \theta = -\frac{1}{2}$ and $\sin \theta = -\frac{\sqrt{3}}{2}$.
Since both $\sin \theta$ and $\cos \theta$ are negative,the complex number lies in the $III$ quadrant.
The reference angle $\alpha$ is given by $\tan \alpha = |\frac{-\sqrt{3}}{-1}| = \sqrt{3}$,so $\alpha = \frac{\pi}{3}$.
In the $III$ quadrant,the argument $\theta = -(\pi - \alpha) = -(\pi - \frac{\pi}{3}) = -\frac{2\pi}{3}$.
Therefore,the modulus is $2$ and the argument is $-\frac{2\pi}{3}$.

(A) Given complex number is $z = -\sqrt{3} + i$.
Let $r \cos \theta = -\sqrt{3}$ and $r \sin \theta = 1$.
On squaring and adding,we get:
$r^{2}(\cos^{2} \theta + \sin^{2} \theta) = (-\sqrt{3})^{2} + (1)^{2}$
$r^{2} = 3 + 1 = 4$
$r = 2$ (since $r > 0$).
Thus,the modulus is $2$.
Now,$2 \cos \theta = -\sqrt{3} \Rightarrow \cos \theta = -\frac{\sqrt{3}}{2}$ and $2 \sin \theta = 1 \Rightarrow \sin \theta = \frac{1}{2}$.
Since $\cos \theta < 0$ and $\sin \theta > 0$,the angle $\theta$ lies in the $II$ quadrant.
The reference angle is $\alpha = \frac{\pi}{6}$.
Therefore,$\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$.
The modulus is $2$ and the argument is $\frac{5\pi}{6}$.

Let the complex number be $z = 1-i$.
We represent $z$ in polar form as $z = r(\cos \theta + i \sin \theta)$,where $r \cos \theta = 1$ and $r \sin \theta = -1$.
Squaring and adding these equations,we get:
$r^2(\cos^2 \theta + \sin^2 \theta) = 1^2 + (-1)^2$
$r^2 = 2$
$r = \sqrt{2}$ (since $r > 0$).
Now,$\cos \theta = \frac{1}{\sqrt{2}}$ and $\sin \theta = -\frac{1}{\sqrt{2}}$.
Since $\cos \theta > 0$ and $\sin \theta < 0$,the angle $\theta$ lies in the $IV$ quadrant.
Thus,$\theta = -\frac{\pi}{4}$.
Therefore,the polar form is $\sqrt{2} \left[ \cos \left( -\frac{\pi}{4} \right) + i \sin \left( -\frac{\pi}{4} \right) \right]$.

(B) Let $z = -1+i$. The polar form is $z = r(\cos \theta + i \sin \theta)$.
Here,$r \cos \theta = -1$ and $r \sin \theta = 1$.
Squaring and adding both equations:
$r^2(\cos^2 \theta + \sin^2 \theta) = (-1)^2 + (1)^2$
$r^2 = 1 + 1 = 2$
$r = \sqrt{2}$ (since $r > 0$).
Now,$\cos \theta = -\frac{1}{\sqrt{2}}$ and $\sin \theta = \frac{1}{\sqrt{2}}$.
Since $\cos \theta < 0$ and $\sin \theta > 0$,$\theta$ lies in the $II$ quadrant.
The reference angle $\alpha$ is given by $\tan \alpha = |\frac{1}{-1}| = 1$,so $\alpha = \frac{\pi}{4}$.
In the $II$ quadrant,$\theta = \pi - \alpha = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.
Thus,the polar form is $\sqrt{2}(\cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4})$.

Let $z = -1-i$.
We represent the complex number in polar form as $z = r(\cos \theta + i \sin \theta)$,where $r = |z| = \sqrt{(-1)^2 + (-1)^2} = \sqrt{1+1} = \sqrt{2}$.
Now,$r \cos \theta = -1$ and $r \sin \theta = -1$.
$\Rightarrow \cos \theta = -\frac{1}{\sqrt{2}}$ and $\sin \theta = -\frac{1}{\sqrt{2}}$.
Since both $\cos \theta$ and $\sin \theta$ are negative,the angle $\theta$ lies in the $III$ quadrant.
The reference angle $\alpha$ is given by $\tan \alpha = |\frac{-1}{-1}| = 1$,so $\alpha = \frac{\pi}{4}$.
In the $III$ quadrant,$\theta = -(\pi - \alpha) = -(\pi - \frac{\pi}{4}) = -\frac{3\pi}{4}$.
Thus,the polar form is $\sqrt{2}(\cos(-\frac{3\pi}{4}) + i \sin(-\frac{3\pi}{4}))$.

(A) Let the complex number be $z = \sqrt{3} + i$.
The polar form is given by $z = r(\cos \theta + i \sin \theta)$,where $r = |z| = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3+1} = \sqrt{4} = 2$.
We have $r \cos \theta = \sqrt{3}$ and $r \sin \theta = 1$.
Substituting $r = 2$,we get $\cos \theta = \frac{\sqrt{3}}{2}$ and $\sin \theta = \frac{1}{2}$.
Since both $\cos \theta$ and $\sin \theta$ are positive,$\theta$ lies in the $I$ quadrant.
The principal argument $\theta = \frac{\pi}{6}$.
Thus,the polar form is $2(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6})$.

(A) Let the complex number be $z = i = 0 + 1i$.
The polar form is given by $z = r(\cos \theta + i \sin \theta)$,where $r \cos \theta = 0$ and $r \sin \theta = 1$.
Squaring and adding both equations:
$r^2(\cos^2 \theta + \sin^2 \theta) = 0^2 + 1^2$
$r^2(1) = 1$
$r = 1$ (since $r > 0$).
Now,$\cos \theta = 0$ and $\sin \theta = 1$.
This implies $\theta = \frac{\pi}{2}$.
Therefore,the polar form is $1(\cos \frac{\pi}{2} + i \sin \frac{\pi}{2}) = \cos \frac{\pi}{2} + i \sin \frac{\pi}{2}$.

(N/A) Given the complex number $z = \frac{1+i}{1-i}$.
Multiply the numerator and denominator by the conjugate of the denominator $(1+i)$:
$z = \frac{1+i}{1-i} \times \frac{1+i}{1+i} = \frac{(1+i)^2}{1^2 - i^2} = \frac{1 + 2i + i^2}{1 - (-1)} = \frac{1 + 2i - 1}{2} = \frac{2i}{2} = i$.
We can write $z = 0 + i$.
Let $z = r(\cos \theta + i \sin \theta)$,where $r$ is the modulus and $\theta$ is the argument.
Comparing $0 + i$ with $r \cos \theta + i r \sin \theta$:
$r \cos \theta = 0$ and $r \sin \theta = 1$.
Squaring and adding: $r^2(\cos^2 \theta + \sin^2 \theta) = 0^2 + 1^2 \implies r^2 = 1 \implies r = 1$ (since $r > 0$).
Now,$\cos \theta = 0$ and $\sin \theta = 1$.
This implies $\theta = \frac{\pi}{2}$.
Thus,the modulus is $1$ and the argument is $\frac{\pi}{2}$.

(A) Let $z = \frac{1+7i}{(2-i)^2}$.
Expanding the denominator: $(2-i)^2 = 4 + i^2 - 4i = 4 - 1 - 4i = 3 - 4i$.
So,$z = \frac{1+7i}{3-4i}$.
Multiply numerator and denominator by the conjugate $(3+4i)$:
$z = \frac{(1+7i)(3+4i)}{(3-4i)(3+4i)} = \frac{3 + 4i + 21i + 28i^2}{3^2 + 4^2} = \frac{3 + 25i - 28}{9 + 16} = \frac{-25 + 25i}{25} = -1 + i$.
For polar form $z = r(\cos \theta + i \sin \theta)$:
$r = \sqrt{(-1)^2 + (1)^2} = \sqrt{1+1} = \sqrt{2}$.
Since $x = -1$ and $y = 1$,the complex number lies in the $II$ quadrant.
$\cos \theta = \frac{-1}{\sqrt{2}}$ and $\sin \theta = \frac{1}{\sqrt{2}}$.
Thus,$\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.
The polar form is $\sqrt{2}\left(\cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4}\right)$.

(A) Given $z = \frac{1+3i}{1-2i}$.
Multiply numerator and denominator by the conjugate $(1+2i)$:
$z = \frac{(1+3i)(1+2i)}{(1-2i)(1+2i)} = \frac{1+2i+3i+6i^2}{1^2+2^2} = \frac{1+5i-6}{5} = \frac{-5+5i}{5} = -1+i$.
Let $z = r(\cos \theta + i \sin \theta)$.
Here,$r = |z| = \sqrt{(-1)^2 + (1)^2} = \sqrt{1+1} = \sqrt{2}$.
Since $z = -1+i$ lies in the second quadrant,$\cos \theta = \frac{-1}{\sqrt{2}}$ and $\sin \theta = \frac{1}{\sqrt{2}}$.
The reference angle is $\alpha = \frac{\pi}{4}$.
Since it is in the second quadrant,$\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.
Thus,the polar form is $\sqrt{2}\left(\cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4}\right)$.

(C) Given $\overline{z}_{1} = i \overline{z}_{2}$. Taking conjugate on both sides,we get $z_{1} = -i z_{2}$.
Substitute $z_{1}$ into the argument equation: $\arg \left( \frac{-i z_{2}}{\overline{z}_{2}} \right) = \pi$.
Using properties of arguments,$\arg(-i) + \arg \left( \frac{z_{2}}{\overline{z}_{2}} \right) = \pi$.
We know $\arg(-i) = -\frac{\pi}{2}$ and $\arg \left( \frac{z_{2}}{\overline{z}_{2}} \right) = \arg(z_{2}) - \arg(\overline{z}_{2}) = \theta - (-\theta) = 2\theta$,where $\theta = \arg(z_{2})$.
So,$-\frac{\pi}{2} + 2\theta = \pi$,which gives $2\theta = \frac{3\pi}{2}$,so $\theta = \frac{3\pi}{4}$.
Now,$z_{1} = -i z_{2} = e^{-i\pi/2} \cdot |z_{2}| e^{i(3\pi/4)} = |z_{2}| e^{i(3\pi/4 - \pi/2)} = |z_{2}| e^{i\pi/4}$.
Thus,$\arg(z_{1}) = \frac{\pi}{4}$.

(D) Given $z = 1 + i$,so $\overline{z} = 1 - i$ and $\frac{1}{z} = \frac{1}{1 + i} = \frac{1 - i}{2}$.
Substitute these into the expression for $z_1$:
$z_1 = \frac{1 + i(1 - i)}{(1 - i)(1 - (1 + i)) + \frac{1 - i}{2}}$
$z_1 = \frac{1 + i - i^2}{(1 - i)(-i) + \frac{1 - i}{2}}$
$z_1 = \frac{1 + i + 1}{-i + i^2 + \frac{1 - i}{2}} = \frac{2 + i}{-i - 1 + \frac{1 - i}{2}}$
$z_1 = \frac{2 + i}{\frac{-2i - 2 + 1 - i}{2}} = \frac{2(2 + i)}{-1 - 3i} = \frac{2(2 + i)}{-(1 + 3i)}$
Multiply numerator and denominator by $(1 - 3i)$:
$z_1 = \frac{-2(2 + i)(1 - 3i)}{1^2 + 3^2} = \frac{-2(2 - 6i + i - 3i^2)}{10} = \frac{-2(2 - 5i + 3)}{10} = \frac{-2(5 - 5i)}{10} = -(1 - i) = -1 + i$.
Now,find $\arg(z_1)$ for $z_1 = -1 + i$:
Since $z_1$ is in the second quadrant,$\arg(z_1) = \pi - \tan^{-1}(1) = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.
Finally,calculate $\frac{12}{\pi} \arg(z_1)$:
$\frac{12}{\pi} \times \frac{3\pi}{4} = 3 \times 3 = 9$.

(A) Given $z = 2 - i(2 \tan \frac{5 \pi}{8})$.
Comparing with $z = x + iy$,we have $x = 2$ and $y = -2 \tan \frac{5 \pi}{8}$.
The modulus $r$ is given by $r = \sqrt{x^2 + y^2} = \sqrt{2^2 + (-2 \tan \frac{5 \pi}{8})^2} = \sqrt{4(1 + \tan^2 \frac{5 \pi}{8})} = \sqrt{4 \sec^2 \frac{5 \pi}{8}} = |2 \sec \frac{5 \pi}{8}|$.
Since $\frac{5 \pi}{8}$ is in the second quadrant,$\sec \frac{5 \pi}{8}$ is negative,so $r = -2 \sec \frac{5 \pi}{8} = 2 \sec(\pi - \frac{5 \pi}{8}) = 2 \sec \frac{3 \pi}{8}$.
The argument $\theta$ is $\tan^{-1}(\frac{y}{x}) = \tan^{-1}(\frac{-2 \tan \frac{5 \pi}{8}}{2}) = \tan^{-1}(-\tan \frac{5 \pi}{8}) = \tan^{-1}(\tan(\pi - \frac{5 \pi}{8})) = \tan^{-1}(\tan \frac{3 \pi}{8}) = \frac{3 \pi}{8}$.
Thus,$(r, \theta) = (2 \sec \frac{3 \pi}{8}, \frac{3 \pi}{8})$.

(A) To find the argument of $z = \frac{13-5i}{4-9i}$,first simplify the complex number by multiplying the numerator and denominator by the conjugate of the denominator,$(4+9i)$:
$z = \frac{(13-5i)(4+9i)}{(4-9i)(4+9i)}$
$z = \frac{52 + 117i - 20i - 45i^2}{16 - 81i^2}$
Since $i^2 = -1$,we have:
$z = \frac{52 + 97i + 45}{16 + 81} = \frac{97 + 97i}{97} = 1 + i$
Now,the complex number is in the form $z = x + iy$ where $x = 1$ and $y = 1$.
The argument $\theta$ is given by $\tan^{-1}(\frac{y}{x})$:
$\theta = \tan^{-1}(\frac{1}{1}) = \tan^{-1}(1) = \frac{\pi}{4}$
Thus,the argument is $\frac{\pi}{4}$.

(A) Given $Z = \frac{-2}{1 + \sqrt{3}i}$.
Rationalizing the denominator:
$Z = \frac{-2(1 - \sqrt{3}i)}{(1 + \sqrt{3}i)(1 - \sqrt{3}i)}$
$Z = \frac{-2(1 - \sqrt{3}i)}{1^2 - (\sqrt{3}i)^2}$
$Z = \frac{-2(1 - \sqrt{3}i)}{1 + 3} = \frac{-2(1 - \sqrt{3}i)}{4}$
$Z = -\frac{1}{2} + \frac{\sqrt{3}}{2}i$.
Here,the real part $a = -\frac{1}{2}$ and the imaginary part $b = \frac{\sqrt{3}}{2}$.
Since $a < 0$ and $b > 0$,the complex number lies in the second quadrant.
$\arg(Z) = \pi - \tan^{-1}\left|\frac{b}{a}\right|$
$\arg(Z) = \pi - \tan^{-1}\left|\frac{\sqrt{3}/2}{-1/2}\right|$
$\arg(Z) = \pi - \tan^{-1}(\sqrt{3})$
$\arg(Z) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.

(A) Given $z_1 = 5 - 2i$ and $z_2 = 3 + i$.
First,calculate $z_1 + z_2 = (5 + 3) + (-2 + 1)i = 8 - i$.
Next,calculate $z_1 - z_2 = (5 - 3) + (-2 - 1)i = 2 - 3i$.
Now,consider the ratio $\frac{z_1 + z_2}{z_1 - z_2} = \frac{8 - i}{2 - 3i}$.
To simplify,multiply the numerator and denominator by the conjugate of the denominator $(2 + 3i)$:
$\frac{8 - i}{2 - 3i} \times \frac{2 + 3i}{2 + 3i} = \frac{16 + 24i - 2i - 3i^2}{2^2 + 3^2} = \frac{16 + 22i + 3}{4 + 9} = \frac{19 + 22i}{13} = \frac{19}{13} + \frac{22}{13}i$.
The argument is given by $\tan^{-1}\left(\frac{\text{imaginary part}}{\text{real part}}\right) = \tan^{-1}\left(\frac{22/13}{19/13}\right) = \tan^{-1}\left(\frac{22}{19}\right)$.

(C) Let $z = \frac{1+i \sqrt{3}}{\sqrt{3}+i}$.
Multiply the numerator and denominator by the conjugate of the denominator $(\sqrt{3}-i)$:
$z = \frac{(1+i \sqrt{3})(\sqrt{3}-i)}{(\sqrt{3}+i)(\sqrt{3}-i)}$
$z = \frac{\sqrt{3} - i + 3i - i^2 \sqrt{3}}{3 - i^2}$
Since $i^2 = -1$,we have:
$z = \frac{\sqrt{3} + 2i + \sqrt{3}}{3 + 1} = \frac{2\sqrt{3} + 2i}{4} = \frac{\sqrt{3}}{2} + \frac{1}{2}i$
The argument of $z = a + bi$ is given by $\theta = \tan^{-1}\left(\frac{b}{a}\right)$.
$\theta = \tan^{-1}\left(\frac{1/2}{\sqrt{3}/2}\right) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$.

(C) We know that $arg(-z) = arg(-1 \times z)$.
Using the property $arg(z_1 z_2) = arg(z_1) + arg(z_2)$,we get $arg(-z) = arg(-1) + arg(z)$.
Since $arg(-1) = \pi$,we have $arg(-z) = \pi + arg(z)$.
Therefore,$arg(-z) - arg(z) = (\pi + arg(z)) - arg(z) = \pi$.

(C) Let $z = \frac{1-i \sqrt{3}}{1+i \sqrt{3}}$.
Multiply the numerator and denominator by the conjugate of the denominator $(1-i \sqrt{3})$:
$z = \frac{(1-i \sqrt{3})(1-i \sqrt{3})}{(1+i \sqrt{3})(1-i \sqrt{3})} = \frac{1 - 2i \sqrt{3} + i^2(3)}{1^2 + (\sqrt{3})^2} = \frac{1 - 2i \sqrt{3} - 3}{1 + 3} = \frac{-2 - 2i \sqrt{3}}{4} = -\frac{1}{2} - i \frac{\sqrt{3}}{2}$.
Since the complex number $z = -\frac{1}{2} - i \frac{\sqrt{3}}{2}$ lies in the third quadrant,the argument is given by $\text{Arg}(z) = -\pi + \tan^{-1}\left|\frac{\text{Im}(z)}{\text{Re}(z)}\right|$ or $\pi + \tan^{-1}\left(\frac{\sqrt{3}/2}{1/2}\right)$.
$\text{Arg}(z) = 180^{\circ} + \tan^{-1}(\sqrt{3}) = 180^{\circ} + 60^{\circ} = 240^{\circ}$.

(D) Let the complex number be $z = r(\cos \theta + i \sin \theta)$.
Given that the distance from the origin is $r = 2$ and the argument is $\theta = \frac{5 \pi}{6}$.
Substituting these values,we get $z = 2 \left( \cos \frac{5 \pi}{6} + i \sin \frac{5 \pi}{6} \right)$.
Since $\cos \frac{5 \pi}{6} = \cos(\pi - \frac{\pi}{6}) = -\cos \frac{\pi}{6} = -\frac{\sqrt{3}}{2}$ and $\sin \frac{5 \pi}{6} = \sin(\pi - \frac{\pi}{6}) = \sin \frac{\pi}{6} = \frac{1}{2}$.
Therefore,$z = 2 \left( -\frac{\sqrt{3}}{2} + i \frac{1}{2} \right) = -\sqrt{3} + i$.

(B) Let $z = (1+i)^{5}$.
First,express $1+i$ in polar form: $1+i = \sqrt{2}(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})$.
Using De Moivre's theorem,$z = (\sqrt{2})^{5}(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})^{5} = 4\sqrt{2}(\cos \frac{5\pi}{4} + i \sin \frac{5\pi}{4})$.
The argument of $z$ is $\frac{5\pi}{4}$.
Since the principal argument must lie in the interval $(-\pi, \pi]$,we subtract $2\pi$ from $\frac{5\pi}{4}$:
$\frac{5\pi}{4} - 2\pi = -\frac{3\pi}{4}$.
Thus,the principal amplitude is $-\frac{3\pi}{4}$.

(B) Let $z = \frac{(1+i \sqrt{3})(-\sqrt{3}-i)}{(1-i)(-i)}$
Numerator: $(1+i \sqrt{3})(-\sqrt{3}-i) = -\sqrt{3} - i - 3i - i^2 \sqrt{3} = -\sqrt{3} - 4i + \sqrt{3} = -4i$
Denominator: $(1-i)(-i) = -i + i^2 = -i - 1 = -(1+i)$
So,$z = \frac{-4i}{-(1+i)} = \frac{4i}{1+i}$
Multiply numerator and denominator by the conjugate $(1-i)$:
$z = \frac{4i(1-i)}{(1+i)(1-i)} = \frac{4i - 4i^2}{1 - i^2} = \frac{4i + 4}{1 + 1} = \frac{4+4i}{2} = 2+2i$
Since $z = 2+2i$ lies in the $I^{st}$ quadrant,the argument is given by $\theta = \tan^{-1}\left(\frac{2}{2}\right) = \tan^{-1}(1) = \frac{\pi}{4}$.

(D) Given $z = \sqrt{\frac{1-i}{1+i}}$.
Multiply numerator and denominator by $(1-i)$:
$z = \sqrt{\frac{(1-i)^2}{1^2+1^2}} = \sqrt{\frac{(1-i)^2}{2}} = \pm \frac{1-i}{\sqrt{2}}$.
Thus,$z_1 = \frac{1-i}{\sqrt{2}}$ and $z_2 = \frac{-1+i}{\sqrt{2}}$.
For $z_1 = \frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}}$,the argument is $\arg(z_1) = \tan^{-1}\left(\frac{-1/\sqrt{2}}{1/\sqrt{2}}\right) = -\frac{\pi}{4}$.
For $z_2 = -\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}$,the argument is $\arg(z_2) = \pi + \tan^{-1}\left(\frac{1/\sqrt{2}}{-1/\sqrt{2}}\right) = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.
Since $-\frac{\pi}{2} < -\frac{\pi}{4} < \frac{3\pi}{4} < \pi$,the condition is satisfied.
Therefore,$\arg(z_1) + \arg(z_2) = -\frac{\pi}{4} + \frac{3\pi}{4} = \frac{2\pi}{4} = \frac{\pi}{2}$.

(B) Given that the amplitude (argument) of the complex number $z = x + iy$ is $\theta$.
By definition,$\arg(z) = \tan^{-1}\left(\frac{y}{x}\right) = \theta$.
Taking the tangent of both sides,we get $\frac{y}{x} = \tan \theta$.
Therefore,the locus of the point is $y = x \tan \theta$,which represents a straight line passing through the origin.

(A) First,simplify the expression inside the argument:
$\frac{4+2 i}{1-2 i} = \frac{(4+2 i)(1+2 i)}{(1-2 i)(1+2 i)} = \frac{4+8 i+2 i-4}{1+4} = \frac{10 i}{5} = 2 i$
$\frac{3+4 i}{2+3 i} = \frac{(3+4 i)(2-3 i)}{(2+3 i)(2-3 i)} = \frac{6-9 i+8 i+12}{4+9} = \frac{18-i}{13}$
Now,add the two results:
$2 i + \frac{18-i}{13} = \frac{26 i + 18 - i}{13} = \frac{18+25 i}{13} = \frac{18}{13} + \frac{25}{13} i$
The argument is $\tan^{-1}\left(\frac{25/13}{18/13}\right) = \tan^{-1}\left(\frac{25}{18}\right)$.
Since $\frac{25}{18} > 1$,we have $\tan^{-1}\left(\frac{25}{18}\right) > \tan^{-1}(1) = \frac{\pi}{4}$.
Also,since $\frac{25}{18} > 0$,the argument lies in the first quadrant,i.e.,$(0, \frac{\pi}{2})$.
Thus,the value lies in the interval $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$.

(A) Given,$\bar{z}+i \bar{w}=0$.
Taking the conjugate on both sides,we get $z-i w=0$,which implies $z=i w$.
We are given $\operatorname{Arg}(z w)=\pi$.
Using the property $\operatorname{Arg}(z w) = \operatorname{Arg}(z) + \operatorname{Arg}(w)$,we have $\operatorname{Arg}(z) + \operatorname{Arg}(w) = \pi$.
Since $z=i w$,we have $w = \frac{z}{i} = -iz$.
Thus,$\operatorname{Arg}(w) = \operatorname{Arg}(-i) + \operatorname{Arg}(z) = -\frac{\pi}{2} + \operatorname{Arg}(z)$.
Substituting this into the equation: $\operatorname{Arg}(z) + (\operatorname{Arg}(z) - \frac{\pi}{2}) = \pi$.
$2 \operatorname{Arg}(z) = \pi + \frac{\pi}{2} = \frac{3 \pi}{2}$.
Therefore,$\operatorname{Arg}(z) = \frac{3 \pi}{4}$.

(C) Given that $\operatorname{Arg} z_1 = \frac{\pi}{3}$.
We know that $\operatorname{Arg} \overline{z_2} = -\operatorname{Arg} z_2$.
Given $\operatorname{Arg} \overline{z_2} = \frac{\pi}{5}$,therefore $\operatorname{Arg} z_2 = -\frac{\pi}{5}$.
Now,$\operatorname{Arg} z_1 + \operatorname{Arg} z_2 = \frac{\pi}{3} - \frac{\pi}{5}$.
Taking the least common multiple,we get $\frac{5\pi - 3\pi}{15} = \frac{2\pi}{15}$.

(C) Given $\bar{z}_1 - i \bar{z}_2 = 0$.
Taking the conjugate on both sides,we get $z_1 + i z_2 = 0$,which implies $z_1 = -i z_2$.
We know that $-i = e^{-i \pi / 2}$,so $z_1 = z_2 e^{-i \pi / 2}$.
Taking the argument on both sides,$\arg(z_1) = \arg(z_2) - \frac{\pi}{2}$,which implies $\arg(z_2) = \arg(z_1) + \frac{\pi}{2}$.
Given $\arg(z_1 z_2) = \arg(z_1) + \arg(z_2) = \frac{3 \pi}{4}$.
Substituting $\arg(z_2)$,we get $\arg(z_1) + (\arg(z_1) + \frac{\pi}{2}) = \frac{3 \pi}{4}$.
$2 \arg(z_1) = \frac{3 \pi}{4} - \frac{\pi}{2} = \frac{\pi}{4}$.
Therefore,$\arg(z_1) = \frac{\pi}{8}$.

(A) Given $\arg \left(\frac{z-1}{z-3i}\right)=\frac{\pi}{2}$.
Let $z=x+iy$. Then $\frac{z-1}{z-3i} = \frac{(x-1)+iy}{x+i(y-3)}$.
Multiplying numerator and denominator by the conjugate of the denominator: $\frac{((x-1)+iy)(x-i(y-3))}{x^2+(y-3)^2} = \frac{x(x-1)+y(y-3) + i(xy - (x-1)(y-3))}{x^2+(y-3)^2}$.
For the argument to be $\frac{\pi}{2}$,the real part must be $0$ and the imaginary part must be positive.
Real part: $x(x-1)+y(y-3)=0 \Rightarrow x^2-x+y^2-3y=0$.
Completing the square: $\left(x-\frac{1}{2}\right)^2 + \left(y-\frac{3}{2}\right)^2 = \frac{1}{4} + \frac{9}{4} = \frac{10}{4} = \left(\frac{\sqrt{10}}{2}\right)^2$.
This represents a circle with center $\left(\frac{1}{2}, \frac{3}{2}\right)$ and radius $\frac{\sqrt{10}}{2}$.
The condition $\arg(w) = \frac{\pi}{2}$ implies the locus is the arc of the circle where the imaginary part is positive.

(B) For statement $I$: $\sqrt{-a} \times \sqrt{-b} = (i\sqrt{a}) \times (i\sqrt{b}) = i^2 \sqrt{ab} = -\sqrt{ab}$. Thus,statement $I$ is false.
For statement $II$: Let $z = \frac{1+i\sqrt{3}}{1-i\sqrt{3}}$. Multiplying numerator and denominator by the conjugate of the denominator $(1+i\sqrt{3})$:
$z = \frac{(1+i\sqrt{3})^2}{1^2 + (\sqrt{3})^2} = \frac{1 - 3 + 2i\sqrt{3}}{4} = \frac{-2 + 2i\sqrt{3}}{4} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$.
This complex number lies in the second quadrant. The argument is $\pi - \tan^{-1}(\frac{\sqrt{3}/2}{1/2}) = 180^{\circ} - 60^{\circ} = 120^{\circ}$. Thus,statement $II$ is true.

(C) Given $z_1 = -\sqrt{3} + i$ and $z_2 = -\sqrt{3} - i$.
We know that the argument of a quotient is given by $\text{arg}(\frac{z_1}{z_2}) = \text{arg}(z_1) - \text{arg}(z_2)$.
For $z_1 = -\sqrt{3} + i$,the point lies in the second quadrant. $\text{arg}(z_1) = \pi - \tan^{-1}(\frac{1}{\sqrt{3}}) = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$.
For $z_2 = -\sqrt{3} - i$,the point lies in the third quadrant. $\text{arg}(z_2) = -(\pi - \tan^{-1}(\frac{1}{\sqrt{3}})) = -(\pi - \frac{\pi}{6}) = -\frac{5\pi}{6}$.
Therefore,$\text{arg}(\frac{z_1}{z_2}) = \frac{5\pi}{6} - (-\frac{5\pi}{6}) = \frac{10\pi}{6} = \frac{5\pi}{3}$.
Since the principal amplitude must lie in the interval $(-\pi, \pi]$,we subtract $2\pi$: $\frac{5\pi}{3} - 2\pi = -\frac{\pi}{3}$.
Thus,the principal amplitude is $-\frac{\pi}{3}$.

(D) $z=1+i \sqrt{3}$
Since $z$ is in the first quadrant,$\operatorname{Arg} z = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3}$.
For the conjugate $\bar{z} = 1-i \sqrt{3}$,which is in the fourth quadrant,$\operatorname{Arg} \bar{z} = \tan^{-1}\left(\frac{-\sqrt{3}}{1}\right) = -\frac{\pi}{3}$.
Therefore,$|\operatorname{Arg} z| + |\operatorname{Arg} \bar{z}| = |\frac{\pi}{3}| + |-\frac{\pi}{3}| = \frac{\pi}{3} + \frac{\pi}{3} = \frac{2 \pi}{3}$.

(B) Let $z = \sin \frac{\pi}{5} + i(1 - \cos \frac{\pi}{5})$.
Using the trigonometric identities $\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$ and $1 - \cos \theta = 2 \sin^2 \frac{\theta}{2}$,we get:
$z = 2 \sin \frac{\pi}{10} \cos \frac{\pi}{10} + i(2 \sin^2 \frac{\pi}{10})$
$z = 2 \sin \frac{\pi}{10} (\cos \frac{\pi}{10} + i \sin \frac{\pi}{10})$
Since $2 \sin \frac{\pi}{10} > 0$,the complex number is in the polar form $r(\cos \theta + i \sin \theta)$ where $r = 2 \sin \frac{\pi}{10}$ and $\theta = \frac{\pi}{10}$.
Thus,the amplitude of the given complex number is $\frac{\pi}{10}$.

(B) For a complex number $z = x + iy$,the polar form is $r \operatorname{cis} \theta$,where $r = \sqrt{x^2 + y^2}$ and $\theta = \operatorname{arg}(z)$.
$(i) z = \sqrt{3} - i$: $r = \sqrt{3+1} = 2$. $\theta = \tan^{-1}(\frac{-1}{\sqrt{3}}) = -\frac{\pi}{6}$. Thus,$z = 2 \operatorname{cis}(-\frac{\pi}{6})$ (Matches $d$).
$(ii) z = \sqrt{3} + i$: $r = 2$. $\theta = \tan^{-1}(\frac{1}{\sqrt{3}}) = \frac{\pi}{6}$. Thus,$z = 2 \operatorname{cis}(\frac{\pi}{6})$ (Matches $a$).
$(iii) z = -\sqrt{3} + i$: $r = 2$. $\theta = \pi - \tan^{-1}(\frac{1}{\sqrt{3}}) = \frac{5\pi}{6}$. Thus,$z = 2 \operatorname{cis}(\frac{5\pi}{6})$ (Matches $b$).
$(iv) z = -\sqrt{3} - i$: $r = 2$. $\theta = -\pi + \tan^{-1}(\frac{1}{\sqrt{3}}) = -\frac{5\pi}{6}$. Thus,$z = 2 \operatorname{cis}(-\frac{5\pi}{6})$ (Matches $c$).
Therefore,the correct matching is $(i)-d, (ii)-a, (iii)-b, (iv)-c$.

(A) Given $z_1 = 2 - i$ and $z_2 = 6 + 3i$.
We need to find $\operatorname{amp}\left(\frac{z_1-z_2}{z_1+z_2}\right)$.
First,calculate $z_1 - z_2 = (2 - 6) + (-1 - 3)i = -4 - 4i$.
Next,calculate $z_1 + z_2 = (2 + 6) + (-1 + 3)i = 8 + 2i$.
Now,$\frac{z_1-z_2}{z_1+z_2} = \frac{-4-4i}{8+2i} = \frac{-2-2i}{4+i}$.
Multiply numerator and denominator by the conjugate $(4-i)$:
$\frac{(-2-2i)(4-i)}{(4+i)(4-i)} = \frac{-8 + 2i - 8i + 2i^2}{16 - i^2} = \frac{-8 - 6i - 2}{16 + 1} = \frac{-10 - 6i}{17} = -\frac{10}{17} - \frac{6}{17}i$.
Alternatively,using the property $\operatorname{amp}\left(\frac{z_1-z_2}{z_1+z_2}\right) = \operatorname{amp}(z_1-z_2) - \operatorname{amp}(z_1+z_2)$:
$\operatorname{amp}(-4-4i) = -\pi + \tan^{-1}\left(\frac{-4}{-4}\right) = -\pi + \frac{\pi}{4} = -\frac{3\pi}{4}$.
$\operatorname{amp}(8+2i) = \tan^{-1}\left(\frac{2}{8}\right) = \tan^{-1}\left(\frac{1}{4}\right)$.
Thus,the result is $-\frac{3\pi}{4} - \tan^{-1}\left(\frac{1}{4}\right)$.

(C) Given $z = x + iy$. Since $|z| = 1$, we have $x^2 + y^2 = 1$.
From $z = 1 - \bar{z}$, we get $x + iy = 1 - (x - iy) = 1 - x + iy$.
Comparing real parts, $x = 1 - x$, which implies $2x = 1$, so $x = \frac{1}{2}$.
Since $x^2 + y^2 = 1$, we have $(\frac{1}{2})^2 + y^2 = 1$, so $y^2 = 1 - \frac{1}{4} = \frac{3}{4}$.
Given $\operatorname{Im}(z) > 0$, we have $y = \frac{\sqrt{3}}{2}$.
Thus, $z = \frac{1}{2} + i\frac{\sqrt{3}}{2}$.
Since $z$ has an imaginary part, Statement-$I$ is false.
The argument of $z$ is $\theta = \tan^{-1}(\frac{\sqrt{3}/2}{1/2}) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$.
Thus, Statement-$II$ is true.

(C) Given $|x|=1$ and $\operatorname{Arg}(x)=2\alpha$,we have $x=e^{i 2\alpha}$.
Given $|y|=1$ and $\operatorname{Arg}(y)=3\beta$,we have $y=e^{i 3\beta}$.
Then $x^6 y^4 = (e^{i 2\alpha})^6 (e^{i 3\beta})^4 = e^{i 12\alpha} e^{i 12\beta} = e^{i 12(\alpha+\beta)}$.
Given $\alpha+\beta = \frac{\pi}{36}$,we substitute this into the expression:
$x^6 y^4 = e^{i 12(\frac{\pi}{36})} = e^{i \frac{\pi}{3}}$.
Now,$x^6 y^4 + \frac{1}{x^6 y^4} = e^{i \frac{\pi}{3}} + e^{-i \frac{\pi}{3}}$.
Using Euler's formula $e^{i \theta} + e^{-i \theta} = 2 \cos \theta$,we get:
$2 \cos(\frac{\pi}{3}) = 2 \times \frac{1}{2} = 1$.

(A) Given $z = \frac{(2-i)(1+i)^3}{(1-i)^2}$.
First,simplify $(1+i)^3 = 1 + 3i + 3i^2 + i^3 = 1 + 3i - 3 - i = -2 + 2i$.
Next,simplify $(1-i)^2 = 1 - 2i + i^2 = 1 - 2i - 1 = -2i$.
Substituting these into $z$,we get $z = \frac{(2-i)(-2+2i)}{-2i} = \frac{(2-i) \cdot 2(-1+i)}{-2i} = \frac{(2-i)(1-i)}{i}$.
Expanding the numerator: $(2-i)(1-i) = 2 - 2i - i + i^2 = 2 - 3i - 1 = 1 - 3i$.
So,$z = \frac{1-3i}{i} = \frac{1-3i}{i} \cdot \frac{-i}{-i} = \frac{-i + 3i^2}{1} = -3 - i$.
Since $z = -3 - i$ lies in the $3^{\text{rd}}$ quadrant,the argument is $\operatorname{Arg}(z) = -\pi + \tan^{-1}\left(\frac{-1}{-3}\right) = -\pi + \tan^{-1}\left(\frac{1}{3}\right)$.

(D) The condition $|Z_1+Z_2|=|Z_1|+|Z_2|$ represents the triangle inequality becoming an equality.
This occurs if and only if the vectors representing $Z_1$ and $Z_2$ in the complex plane are in the same direction.
Therefore,the arguments (amplitudes) of $Z_1$ and $Z_2$ must be equal,i.e.,$\text{arg}(Z_1) = \text{arg}(Z_2)$.
Thus,the difference in their amplitudes is $\text{arg}(Z_1) - \text{arg}(Z_2) = 0$.

(D) Let $Z = \sin \frac{6 \pi}{5} + i(1 + \cos \frac{6 \pi}{5})$.
Using trigonometric identities $\sin 2\theta = 2\sin\theta\cos\theta$ and $1 + \cos 2\theta = 2\cos^2\theta$ with $\theta = \frac{3\pi}{5}$:
$Z = 2\sin \frac{3\pi}{5}\cos \frac{3\pi}{5} + i(2\cos^2 \frac{3\pi}{5})$
$Z = 2\cos \frac{3\pi}{5} (\sin \frac{3\pi}{5} + i\cos \frac{3\pi}{5})$
Since $\frac{3\pi}{5} = \frac{\pi}{2} + \frac{\pi}{10}$,we have $\sin \frac{3\pi}{5} = \cos \frac{\pi}{10}$ and $\cos \frac{3\pi}{5} = -\sin \frac{\pi}{10}$.
$Z = 2\cos \frac{3\pi}{5} (\cos \frac{\pi}{10} - i\sin \frac{\pi}{10})$
$Z = 2\cos \frac{3\pi}{5} (\cos(-\frac{\pi}{10}) + i\sin(-\frac{\pi}{10}))$
Since $\cos \frac{3\pi}{5} < 0$,the argument is $\pi - \frac{\pi}{10} = \frac{9\pi}{10}$.

(C) Given the equation $z^2-i=0$,we have $z^2=i$.
Expressing $i$ in polar form,$i = \cos(\frac{\pi}{2}) + i \sin(\frac{\pi}{2}) = e^{i\frac{\pi}{2}}$.
The roots are $z = \pm e^{i\frac{\pi}{4}}$.
Thus,the roots are $z_1 = e^{i\frac{\pi}{4}}$ and $z_2 = e^{i(\frac{\pi}{4} + \pi)} = e^{i\frac{5\pi}{4}}$.
Let $\alpha = e^{i\frac{\pi}{4}}$ and $\beta = e^{i\frac{5\pi}{4}}$.
Then $\operatorname{Arg} \alpha = \frac{\pi}{4}$ and $\operatorname{Arg} \beta = \frac{5\pi}{4}$.
Therefore,$|\operatorname{Arg} \beta - \operatorname{Arg} \alpha| = |\frac{5\pi}{4} - \frac{\pi}{4}| = |\pi| = \pi$.

(A) Let $Z = \frac{(1+i)^{2025}}{(1-i)^{2022}}$.
We know that $1+i = \sqrt{2} e^{i\pi/4}$ and $1-i = \sqrt{2} e^{-i\pi/4}$.
Substituting these values:
$Z = \frac{(\sqrt{2} e^{i\pi/4})^{2025}}{(\sqrt{2} e^{-i\pi/4})^{2022}}$
$Z = \frac{(\sqrt{2})^{2025} e^{i(2025\pi/4)}}{(\sqrt{2})^{2022} e^{-i(2022\pi/4)}}$
$Z = (\sqrt{2})^3 e^{i(2025\pi/4 + 2022\pi/4)}$
$Z = 2\sqrt{2} e^{i(4047\pi/4)}$
Since $4047\pi/4 = 1011\pi + 3\pi/4$,the principal argument is $\operatorname{Arg}(Z) = \frac{3\pi}{4} - \pi = -\frac{\pi}{4}$.

(C) Given,$\arg(\bar{z}_1) = \frac{\pi}{5}$ and $\arg(z_2) = \frac{\pi}{3}$.
We know that $\arg(\bar{z}_1) = -\arg(z_1)$,so $\arg(z_1) = -\frac{\pi}{5}$.
Then,$\arg(z_1 z_2) = \arg(z_1) + \arg(z_2) = -\frac{\pi}{5} + \frac{\pi}{3} = \frac{-3\pi + 5\pi}{15} = \frac{2\pi}{15}$.
Thus,Assertion $(A)$ is true.
For Reason $(R)$,we know that $\arg(\bar{z}) = -\arg(z)$,not $\frac{\pi}{2} + \arg(z)$.
Therefore,Reason $(R)$ is false.
Hence,$(A)$ is true but $(R)$ is false.

(B) Given $\bar{z} - i \bar{w} = 0$,we have $\bar{z} = i \bar{w}$.
Taking the conjugate on both sides,$z = -i w$,which implies $w = \frac{z}{-i} = iz$.
Now,$\operatorname{Arg}(zw) = \operatorname{Arg}(z(iz)) = \operatorname{Arg}(iz^2) = \frac{3 \pi}{4}$.
Using the property $\operatorname{Arg}(z_1 z_2) = \operatorname{Arg}(z_1) + \operatorname{Arg}(z_2)$,we get $\operatorname{Arg}(i) + \operatorname{Arg}(z^2) = \frac{3 \pi}{4}$.
Since $\operatorname{Arg}(i) = \frac{\pi}{2}$ and $\operatorname{Arg}(z^2) = 2 \operatorname{Arg}(z)$,we have $\frac{\pi}{2} + 2 \operatorname{Arg}(z) = \frac{3 \pi}{4}$.
$2 \operatorname{Arg}(z) = \frac{3 \pi}{4} - \frac{\pi}{2} = \frac{\pi}{4}$.
Therefore,$\operatorname{Arg}(z) = \frac{\pi}{8}$.

(B) Given $z = \sqrt{2} \sqrt{1 + \sqrt{3} i}$.
We can write $1 + \sqrt{3} i = \frac{1}{2} (2 + 2 \sqrt{3} i) = \frac{1}{2} ((\sqrt{3})^2 + i^2 + 2 \sqrt{3} i) = \frac{1}{2} (\sqrt{3} + i)^2$.
Thus,$z = \sqrt{2} \cdot \frac{1}{\sqrt{2}} (\sqrt{3} + i) = \pm (\sqrt{3} + i)$.
Since $z$ lies in the third quadrant,both real and imaginary parts must be negative.
Therefore,$z = -\sqrt{3} - i$.
The modulus is $|z| = \sqrt{(-\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = 2$.
The argument $\theta$ in the third quadrant is given by $\theta = -(\pi - \tan^{-1}(\frac{1}{\sqrt{3}})) = -(\pi - \frac{\pi}{6}) = -\frac{5 \pi}{6}$.
Thus,the polar form is $z = 2 \left[ \cos \left( \frac{-5 \pi}{6} \right) + i \sin \left( \frac{-5 \pi}{6} \right) \right]$.

(A) Let $z = x + iy$. Then $z - 1 - 2i = (x - 1) + i(y - 2)$.
Given that $\text{arg}(z - 1 - 2i) = \frac{\pi}{3}$.
This implies $\tan\left(\frac{\pi}{3}\right) = \frac{y - 2}{x - 1}$,where $x > 1$ and $y > 2$.
Since $\tan\left(\frac{\pi}{3}\right) = \sqrt{3}$,we have $\sqrt{3} = \frac{y - 2}{x - 1}$.
Rearranging the terms,we get $y - 2 = \sqrt{3}(x - 1)$.
$y = \sqrt{3}x - \sqrt{3} + 2$.
$y = \sqrt{3}x + (2 - \sqrt{3})$.

(D) Let $z = \frac{(\sqrt{3}+i)(1-\sqrt{3} i)}{(-1+i)(-1-i)}$.
First,simplify the numerator: $(\sqrt{3}+i)(1-\sqrt{3} i) = \sqrt{3} - 3i + i - \sqrt{3} i^2 = \sqrt{3} - 2i + \sqrt{3} = 2\sqrt{3} - 2i$.
Next,simplify the denominator: $(-1+i)(-1-i) = (-1)^2 - (i)^2 = 1 - (-1) = 2$.
Thus,$z = \frac{2\sqrt{3} - 2i}{2} = \sqrt{3} - i$.
The amplitude $\theta$ is given by $\tan^{-1}(\frac{y}{x}) = \tan^{-1}(\frac{-1}{\sqrt{3}})$.
Since the complex number lies in the fourth quadrant $(x > 0, y < 0)$,the amplitude is $-\frac{\pi}{6}$.

(A) The argument of a product of complex numbers is given by $\arg(z_1 z_2) = \arg z_1 + \arg z_2 + 2k\pi$,where $k \in \{0, 1, -1\}$.
Since the principal value of the argument lies in the interval $(-\pi, \pi]$,the sum of the principal arguments may fall outside this range.
Therefore,the principal value of $\arg(z_1 z_2)$ is not necessarily equal to the sum of the principal values of $\arg z_1$ and $\arg z_2$.
Similarly,for the quotient,$\arg(z_1 / z_2) = \arg z_1 - \arg z_2 + 2k\pi$,which also may not equal the difference of the principal values.