Math, asked by fatimaeashal13, 3 months ago

Q. find a+ib

3-i/ 2+i + 3+i/2-i​

Answers

Answered by safdersana80
2

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.

Similar questions