Amplitude of a Complex Number

If $\arg \left(z\right)<0,$ then $\arg \left(-z\right)-\arg \left(z\right)=$

Find the argument of a conjugate of a complex number $z=3+2i$.

A complex number $z=1+i\sqrt{3}$.

The general argument of $z$ is

A complex number $z=\sqrt{3}+i$.

The general argument of $z$ is $2n\mathrm{\pi }+\frac{\mathrm{\pi }}{6}$, where $n$ is an integer.

A complex number $z=1-i$.

The argument of $\overline{z}$ is

A complex number $z=3+9i$.

Find the argument of $\overline{z}$.

The locus of a point $P\left(z\right)$ satisfying $|z+3|+|z –3|=10$ is (where $z$ is a complex number)

The modulus-amplitude form of $\frac{(1-i{)}^3(2-i)}{(2+i)(1+i)}$ is

Statement I  Both $z_1$ and $z_2$ are purely real , if  $arg (z_1 z_2) = 2\pi$  ($z_1$ and $z_2$ have principle arguments).
Statement II Principle arguments of complex number lies between $(-\pi , \pi ].$

Let z= $\frac{(2\sqrt{3}+2i{)}^8}{(1-i{)}^6}+\frac{(1+i{)}^6}{(2\sqrt{3}-2i{)}^8}$. Let θ be the argument of z such that θ∈ (– π,π] then 4 sinθ is equal to

If $\left|\begin{array}{ccc}p& q& r\\ q& r& p\\ r& p& q\end{array}\right|$=0, where p, q, r all the moduli of non-zero complex numbers z1, z2, z3, then prove that arg $\left(\frac{z_3}{z_2}\right)$=λ arg $\left(\frac{z_3-z_1}{z_2-z_1}\right)$ find λ

Consider a square $OABC$ in argand plane, where $O$ is origin and $A$ be complex number $z_0$. Then the equation of the circle that can be inscribed in this square is (Vertices of square are given in anticlockwise order and $i=\sqrt{-1}$)

The argument of $\frac{1+i\sqrt{3}}{\sqrt{3}+1}$ is equal to

If $\left|z-25i\right|\le 15$ , then $|$Maximum $arg\left(z\right)-$Minimum $arg\left(z\right)|$ equals -

If P and Q are represented by the complex numbers $z_1$ and $z_2$, such that $\left|\frac{1}{z_2}+\frac{1}{z_1}\right|= \left|\frac{1}{z_2}-\frac{1}{z_1}\right|$, then the circumcenter of $\mathrm{\Delta }OPQ$ (where O is the origin) is

If $\arg \left({\overline{z}}_1\right)=\arg \left(z_2\right), \left(z\ne 0\right)$ then

If $z_1 \& z_2$ satisfy $z+\overline{z}=2\left|z-1\right|$ and $\arg \left(z_1-z_2\right)=\frac{\pi }{4},$ then $\mathrm{Im}\left(z_1+z_2\right)=$

If ${\mathrm{x}}_{\mathrm{r}}=\cos \frac{\mathrm{\pi }}{2^{\mathrm{r}}}+\mathrm{i}\sin \frac{\mathrm{\pi }}{2^{\mathrm{r}}},\mathrm{ }{\mathrm{z}}_{\mathrm{t}}=\cos \frac{\mathrm{\pi }}{3^{\mathrm{t}}}+\mathrm{i}\sin \frac{\mathrm{\pi }}{3^{\mathrm{t}}}\mathrm{r}=1,\mathrm{ }2,\mathrm{ }3,\mathrm{ }\dots .,\mathrm{ }\mathrm{t}=1,\mathrm{ }2,\mathrm{ }3,\mathrm{ }\dots .\mathrm{ }$ The value of ${\left({\mathrm{x}}_1{\mathrm{x}}_2{\mathrm{x}}_3\dots ..\mathrm{ }\mathrm{\infty }\right)}^2{\left({\mathrm{z}}_1{\mathrm{z}}_2{\mathrm{z}}_3\dots \dots \mathrm{\infty }\right)}^4$ -

If $\left(\frac{3-{\mathrm{z}}_1}{2-{\mathrm{z}}_1}\right)\left(\frac{2-{\mathrm{z}}_2}{3-{\mathrm{z}}_2}\right)=\mathrm{k}$ , then points $\mathrm{A}\left({\mathrm{z}}_1\right),\mathrm{ }\mathrm{B}\left({\mathrm{z}}_2\right),\mathrm{ }\mathrm{C}\left(3,\mathrm{ }0\right)$ and $\mathrm{D}\left(2,\mathrm{ }0\right)$ taken clockwise -

If $\left(\frac{3-{\mathrm{z}}_1}{2-{\mathrm{z}}_1}\right)\left(\frac{2-{\mathrm{z}}_2}{3-{\mathrm{z}}_2}\right)=\mathrm{k}$ , then points $\mathrm{A}\left({\mathrm{z}}_1\right),\mathrm{ }\mathrm{B}\left({\mathrm{z}}_2\right),\mathrm{ }\mathrm{C}\left(3,\mathrm{ }0\right)$ and $\mathrm{D}\mathrm{ }\left(2,\mathrm{ }0\right)$ (taken in clockwise sense) will