Question

Solve the following and find the value of $\dfrac{2+5i}{3-2i}+\dfrac{2-5i}{3+2i}=$

Answer 1

We are given to solve the complex numbers. In order to add the given complex numbers, as usual we will be taking the LCM of the denominators and then we will be multiplying them with the numerators and solving them. Solving the numerators and the denominators will give us the required answer.

Complete step by step solution:
Now let us learn about the complex numbers. Complex numbers are nothing but those numbers that contain real number parts as well as the imaginary number. A complex number is generally written in the form of $a+bi$. The either part of the number can be zero. All of the complex numbers are considered either positive or negative. The imaginary part in the complex number is said to satisfy the condition ${{i}^{2}}=-1$. Also the sum of two conjugate complex numbers is real.
Now let us add the given complex numbers i.e. $\dfrac{2+5i}{3-2i}+\dfrac{2-5i}{3+2i}=$
In order to add them, firstly we have to take the LCM of the denominators.
The LCM would be $\left( 3-2i \right)\left( 3+2i \right)$
Now upon multiplying them to the numerators, we get
$\Rightarrow \dfrac{\left( 2+5i \right)\left( 3+2i \right)+\left( 2-5i \right)\left( 3-2i \right)}{\left( 3-2i \right)\left( 3+2i \right)}$
Upon solving them, we get
$\Rightarrow \dfrac{\left( 6+4i+15i+10{{i}^{2}} \right)+\left( 6-4i-15i+10{{i}^{2}} \right)}{\left( {{3}^{2}}-4{{i}^{2}} \right)}$
Now we will be adding the like terms and we will be substituting ${{i}^{2}}=-1$. Then we get,

& \Rightarrow \dfrac{\left( -4+19i-4-19i \right)}{13} \\\ & \Rightarrow \dfrac{-8}{13} \\\ \end{aligned}$$ $$\therefore $$ $$\dfrac{2+5i}{3-2i}+\dfrac{2-5i}{3+2i}=$$$$\dfrac{-8}{13}$$ **Note:** For easy evaluation of the solution, we have to substitute the value $${{i}^{2}}=-1$$, if the term is present. We must have a note that if the complex number we have is $$a+bi$$ where $$a,b$$ are real numbers and $$a+bi=0$$ then $$a,b$$ will be equal to zero.