Question

((3+2i)/(2-3i))+((3-2i)/(2+3i)) prove this complex number purely real number

Answer 1

(3+2i) (3-2i) 3(3-2i)-2i(3-2i)
-------- + -------= - - - - - - - - - - -
(2-3i) (2+3i) 2(2+3i)-3i(2+3i)

(9-6i) -(6i+4i^2) ( 9 - 6 I) -(6i-4)
= - - - - - - - - - - - = - - - - - - - - -- -
(4+6i)-(6i -9i^2) (4+6i)-(6i +9)
9-6i-6i-4 5
= -------------= -----//
4+6i-6i+9 13

Answer 2

Answer: It becomes a purely real number.

Step-by-step explanation:

Since we have given that

$\dfrac{3+2i}{2-3i}+\dfrac{3-2i}{2+3i}$

We need to prove this is a purely real number.

First we rationalise the denominator:

$\dfrac{3+2i}{2-3i}\times \dfrac{2+3i}{2+3i}\\\\=\dfrac{6+9i+4i-6}{4+9}\\\\=\dfrac{13i}{13}\\\\=i$

Similarly,

$\dfrac{3-2i}{2+3i}\times \dfrac{2-3i}{2-3i}\\\\=\dfrac{6-9i-4i-6}{4+9}\\\\=\dfrac{-13i}{13}\\\\=-i$

So, it becomes,

$i-i=0$

Hence, it becomes a purely real number.