Find the conjugate of $\frac{(3-2 i)(2+3 i)}{(1+2 i)(2-i)}$.
Find the conjugate of $\frac{(3-2 i)(2+3 i)}{(1+2 i)(2-i)}$.
We have, $\frac{(3-2 i)(2+3 i)}{(1+2 i)(2-i)}$
$=\frac{6+9 i-4 i+6}{2-i+4 i+2}=\frac{12+5 i}{4+3 i} \times \frac{4-3 i}{4-3 i} $
$=\frac{48-36 i+20 i+15}{16+9}=\frac{63-16 i}{25}=\frac{63}{25}-\frac{16}{25} i$
Therefore, conjugate of $\frac{(3-2 i)(2+3 i)}{(1+2 i)(2-i)}$ is $\frac{63}{25}+\frac{16}{25} i$.
Similar Questions
Let $z$ be a complex number (not lying on $X$-axis) of maximum modulus such that $\left| {z + \frac{1}{z}} \right| = 1$. Then
Let $z$ be a complex number (not lying on $X$-axis) of maximum modulus such that $\left| {z + \frac{1}{z}} \right| = 1$. Then
For any two complex numbers ${z_1},{z_2}$we have $|{z_1} + {z_2}{|^2} = $ $|{z_1}{|^2} + |{z_2}{|^2}$ then
For any two complex numbers ${z_1},{z_2}$we have $|{z_1} + {z_2}{|^2} = $ $|{z_1}{|^2} + |{z_2}{|^2}$ then
If the conjugate of $(x + iy)(1 - 2i)$ be $1 + i$, then
If the conjugate of $(x + iy)(1 - 2i)$ be $1 + i$, then
If $z$ is a complex number such that ${z^2} = {(\bar z)^2},$ then
If $z$ is a complex number such that ${z^2} = {(\bar z)^2},$ then
If ${z_1},{z_2},{z_3}$be three non-zero complex number, such that ${z_2} \ne {z_1},a = |{z_1}|,b = |{z_2}|$ and $c = |{z_3}|$ suppose that $\left| {\begin{array}{*{20}{c}}a&b&c\\b&c&a\\c&a&b\end{array}} \right| = 0$, then $arg\left( {\frac{{{z_3}}}{{{z_2}}}} \right)$ is equal to
If ${z_1},{z_2},{z_3}$be three non-zero complex number, such that ${z_2} \ne {z_1},a = |{z_1}|,b = |{z_2}|$ and $c = |{z_3}|$ suppose that $\left| {\begin{array}{*{20}{c}}a&b&c\\b&c&a\\c&a&b\end{array}} \right| = 0$, then $arg\left( {\frac{{{z_3}}}{{{z_2}}}} \right)$ is equal to