Question

Write the conjugate of (3-i)²/2+i

Answer 1

Answer:

$\frac{ {3}^{2} + {i}^{2} - 2(3)(i) }{2 + i} \\ \frac{9 + ( - 1) - 6i}{2 + i} \\ \frac{8 - 6i}{2 + i} \times \frac{2 - i}{2 - i} \\ \frac{8(2 - i) - 6i(2 - i)}{ {2}^{2} - {i}^{2} } \\ \frac{16 - 8i - 12i + 6 {i}^{2} }{4 - ( - 1)} \\ \frac{16 - 20i - 6}{5} \\ \frac{10 - 20i}{5} \\ \frac{5(2 - 4i)}{5} \\ 2 - 4i \\ 2(1 - 2i)$

Answer 2

Step-by-step explanation:

It is Given That ,

$z = \frac{ {(3 - i)}^{2} }{2 + i}$

now By using Identity,

${(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab$

We Get,

$\frac{ ({3}^{2} + {i}^{2} - 2 \times 3 \times i )}{2 + i}$

We Know that,

${i}^{2} = - 1$

putting value,

we get,

$\frac{9 - 1 - 6i}{2 + i}$

$z = \frac{8 - 6i}{2 + i}$

Now , Rationalize it,

$\frac{8 - 6i}{2 + i} \times \frac{2 - i}{2 - i}$

$z = \frac{(8 - 6i)(2 - i)}{(2 + i)(2 - i)}$

$z = \frac{16 - 8i - 12i + 6 {i}^{2} }{ {2}^{2} - {i}^{2} }$

Again Filling The Value of

${i}^{2} = - 1$

We get,

$z = \frac{16 -20i - 6}{4 - ( - 1)}$

$z = \frac{10 - 20i}{4 + 1}$

$z = \frac{10 - 20i}{5}$

Split Both Imaginary and Real Roots,

$z = \frac{10}{5} - \frac{20i}{5}$

$z = 2 - 4i$

Now We Have To Find the Conjugate ,

Conjugate Means To Simply Change the Signs of imaginary Root,

Conjugate is equal To ,

$\frac{}{z} = 2 + 4i$

Hope it Helpsss You....