Let $z$ be complex number such that $\left|\frac{z-i}{z+2 i}\right|=1$ and $|z|=\frac{5}{2} \cdot$ Then the value of $|z+3 i|$ is
$\sqrt{10}$
$2 \sqrt{3}$
$\frac{7}{2}$
$\frac{15}{4}$
If complex numbers $z_1$, $z_2$ are such that $\left| {{z_1}} \right| = \sqrt 2 ,\left| {{z_2}} \right| = \sqrt 3$ and $\left| {{z_1} + {z_2}} \right| = \sqrt {5 - 2\sqrt 3 }$, then the value of $|Arg z_1 -Arg z_2|$ is
Consider the following two statements :
Statement $I$ : For any two non-zero complex numbers $\mathrm{z}_1, \mathrm{z}_2$
$\left(\left|z_1\right|+\left|z_2\right|\right)\left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right| \leq 2\left(\left|z_1\right|+\left|z_2\right|\right)$ and
Statement $II$ : If $\mathrm{x}, \mathrm{y}, \mathrm{z}$ are three distinct complex numbers and a, b, c are three positive real numbers such that $\frac{a}{|y-z|}=\frac{b}{|z-x|}=\frac{c}{|x-y|}$, then
$\frac{\mathrm{a}^2}{\mathrm{y}-\mathrm{z}}+\frac{\mathrm{b}^2}{\mathrm{z}-\mathrm{x}}+\frac{\mathrm{c}^2}{\mathrm{x}-\mathrm{y}}=1$
Between the above two statements,
If ${(\sqrt 8 + i)^{50}} = {3^{49}}(a + ib)$ then ${a^2} + {b^2}$ is
Let $z_k=\cos \left(\frac{2 k \pi}{10}\right)+ i \sin \left(\frac{2 k \pi}{10}\right) ; k =1,2, \ldots 9$.
List $I$ | List $II$ |
$P.$ For each $z_k$ there exists a $z_j$ such that $z_k \cdot z_j=1$ | $1.$ True |
$Q.$ There exists a $k \in\{1,2, \ldots ., 9\}$ such that $z_{1 .} . z=z_k$ has no solution $z$ in the set of complex numbers. | $2.$ False |
$R.$ $\frac{\left|1-z_1\right|\left|1-z_2\right| \ldots . .\left|1-z_9\right|}{10}$ equals | $3.$ $1$ |
$S.$ $1-\sum_{k=1}^9 \cos \left(\frac{2 k \pi}{10}\right)$ equals | $4.$ $2$ |
Codes: $ \quad P \quad Q \quad R \quad S$
Find the conjugate of $\frac{(3-2 i)(2+3 i)}{(1+2 i)(2-i)}$.