Question

Q. find a+ib

3-i/ 2+i + 3+i/2-i​

Answer 1

Answer:

Take L.C.M

=

$= \frac{3 - i}{2 + i} + \frac{3 + i}{2 - i}$

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

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

$= \frac{6 - 3i - 2i + {i}^{2} + 6 + 3i + 2i + {i}^{2} }{4 - {i}^{2} }$

By arranging terms:

$= \frac{6 + 6 + {i}^{2} + {i}^{2} }{4 - {i}^{2} }$

$= \frac{12 + 2 {i}^{2} }{4 - {i}^{2} }$

As we know that the value of " i "

$= \frac{12 + 2( - 1)}{4 - ( - 1)}$

$= \frac{12 - 2}{4 + 1}$

$= \frac{10}{5}$

$= 2$

So in the shape of a+bi we can write as:

$= 2 + 0i$

which is the required answer.