Q.Write in polar form of the following complex numbers

Since the argument of $0$ requires a well-defined direction for a vector of zero length, which does not exist, $\arg(0)$ is undefined.

1. For a nonzero complex number $z=x+iy\ne0$, the argument $\theta=\arg(z)$ is the angle made by the vector $(x,y)$ with the positive real axis, found from $x=r\cos\theta$, $y=r\sin\theta$ where $r=|z|=\sqrt{x^2+y^2}>0$.
2. For $z=0$, we have $x=0,\ y=0$, so $r=|z|=0$.
3. The equations $0=0\cdot\cos\theta$ and $0=0\cdot\sin\theta$ are satisfied by EVERY value of $\theta$, since both sides are $0$ regardless of $\theta$.
4. Because no single value of $\theta$ is singled out (every angle works equally), there is no well-defined direction associated with the zero vector, so the argument cannot be assigned any specific value.
5. By convention in complex analysis, $\arg(0)$ is therefore left undefined (it is not $0$, not $\pi$, and not $\infty$).

Writing $e^{3-i\pi/4}$ in polar form and cubing via De Moivre's theorem gives modulus $e^9$ and amplitude $-3\pi/4$.

1. Write $e^{3-i\pi/4} = e^3 \cdot e^{-i\pi/4} = e^3\left(\cos\dfrac{\pi}{4} - i\sin\dfrac{\pi}{4}\right)$.
2. This complex number has modulus $r=e^3$ and amplitude $\theta=-\dfrac{\pi}{4}$.
3. Raising a complex number in polar form $r(\cos\theta+i\sin\theta)$ to the power 3 gives, by De Moivre's theorem, $r^3(\cos 3\theta + i\sin 3\theta)$.
4. New modulus $= r^3 = (e^3)^3 = e^9$.
5. New amplitude $= 3\theta = 3\left(-\dfrac{\pi}{4}\right) = -\dfrac{3\pi}{4}$.
6. So $\left[e^{3-i\pi/4}\right]^3 = e^9\left(\cos\dfrac{-3\pi}{4}+i\sin\dfrac{-3\pi}{4}\right)$, i.e. modulus $e^9$ and amplitude $-\dfrac{3\pi}{4}$.

Principal argument is conventionally taken in $(-\pi,\pi]$.

1. For a complex number $z=x+iy\neq0$, $\arg z$ can take infinitely many values differing from each other by multiples of $2\pi$.
2. To make it single-valued, the principal value $\mathrm{Arg}(z)$ is restricted to one interval of length $2\pi$.
3. The standard convention (used in the TN board syllabus) chooses the half-open interval $(-\pi,\pi]$, i.e. $-\pi < \mathrm{Arg}(z) \le \pi$.
4. Options (a) and (c) only span half the plane, and (d) excludes all positive angles — none give the full single-valued range that the principal value convention requires.