Given:
[tex]$\frac{2 i}{2+i}-\frac{3 i}{3+i}=a+b i[/tex]
To find:
The value of a and b.
Solution:
[tex]$\frac{2 i}{2+i}-\frac{3 i}{3+i}[/tex]
LCM of [tex]2+i, 3+i=(2+i)(3+i)[/tex]
Make the denominator as LCM.
[tex]$=\frac{2 i(3+i)}{(2+i)(3+i)}-\frac{3 i(2+i)}{(3+i)(2+i)}[/tex]
[tex]$=\frac{2 i(3+i)-3 i(2+i)}{(2+i)(3+i)}[/tex]
Multiply the common term into inside the bracket.
[tex]$=\frac{6i+2i^2-6i-3i^2}{6+2i+3i+i^2}[/tex]
The value of i² = -1
[tex]$=\frac{6i-2-6i+3}{6+2i+3i-1}[/tex]
[tex]$=\frac{1}{5+5i}[/tex]
Rationalize the denominator:
Multiply the conjugate.
[tex]$=\frac{1}{5+5i}\times \frac{5-5i}{5-5i}[/tex]
[tex]$=\frac{5-5i}{5^2-(5i)^2}[/tex]
[tex]$=\frac{5-5i}{50}[/tex]
[tex]$=\frac{5}{50}-\frac{5i}{50}[/tex]
Cancelling the common factors, we get
[tex]$=\frac{1}{10}-\frac{1}{10} i[/tex]
The value of a is [tex]\frac{1}{10}[/tex] and the value of b is[tex]-\frac{1}{10}[/tex].
