Answer:
Option A is correct.
Step-by-step explanation:
We are given:
[tex]\frac{2i}{2+i}-\frac{3i}{3+i} = a+bi[/tex]
We need to find the value of a.
The LCM of (2+i) and (3+i) is (2+i)(3+i)
[tex]=\frac{2i(3+i)}{(2+i)(3+i)}-\frac{3i(2+i)}{(2+i)(3+i)}\\=\frac{6i+2i^2}{(2+i)(3+i)}-\frac{6i+3i^2}{(2+i)(3+i)}\\=\frac{6i+2i^2-(6i+3i^2)}{(2+i)(3+i)}\\=\frac{6i+2i^2-6i-3i^2)}{5+5i}\\=\frac{-i^2}{5+5i}\\i^2=-1\\=\frac{-(-1)}{5+5i}\\=\frac{1}{5+5i}[/tex]
Now rationalize the denominator by multiplying by 5-5i/5-5i
[tex]=\frac{1}{5+5i}*\frac{5-5i}{5-5i} \\=\frac{5-5i}{(5+5i)(5-5i)}\\=\frac{5-5i}{(5+5i)(5-5i)}\\(a+b)(a-b)= a^2-b^2\\=\frac{5(1-i)}{(5)^2-(5i)^2}\\=\frac{5(1-i)}{25+25}\\=\frac{5(1-i)}{50}\\=\frac{1-i}{10}\\=\frac{1}{10}-\frac{i}{10}[/tex]
We are given
[tex]\frac{2i}{2+i}-\frac{3i}{3+i} = a+bi[/tex]
Now after solving we have:
[tex]\frac{1}{10}-\frac{i}{10}=a+bi[/tex]
So value of a = 1/10 and value of b = -1/10
So, Option A is correct.
