Question

If (2i/2+i)-(3i/3+i)=a+bi, then a=A. 1/10B. -10C. 1/50D. -1/10​

Answer 1

Answer:

Option A is correct.

Step-by-step explanation:

We are given:

$\frac{2i}{2+i}-\frac{3i}{3+i} = a+bi$

We need to find the value of a.

The LCM of (2+i) and (3+i)  is (2+i)(3+i)

$=\frac{2i(3+i)}{(2+i)(3+i)}-\frac{3i(2+i)}{(2+i)(3+i)}\\=\frac{6i+2i^2}{(2+i)(3+i)}-\frac{6i+3i^2}{(2+i)(3+i)}\\=\frac{6i+2i^2-(6i+3i^2)}{(2+i)(3+i)}\\=\frac{6i+2i^2-6i-3i^2)}{5+5i}\\=\frac{-i^2}{5+5i}\\i^2=-1\\=\frac{-(-1)}{5+5i}\\=\frac{1}{5+5i}$

Now rationalize the denominator by multiplying by 5-5i/5-5i

$=\frac{1}{5+5i}*\frac{5-5i}{5-5i} \\=\frac{5-5i}{(5+5i)(5-5i)}\\=\frac{5-5i}{(5+5i)(5-5i)}\\(a+b)(a-b)= a^2-b^2\\=\frac{5(1-i)}{(5)^2-(5i)^2}\\=\frac{5(1-i)}{25+25}\\=\frac{5(1-i)}{50}\\=\frac{1-i}{10}\\=\frac{1}{10}-\frac{i}{10}$

We are given

$\frac{2i}{2+i}-\frac{3i}{3+i} = a+bi$

Now after solving we have:

$\frac{1}{10}-\frac{i}{10}=a+bi$

So value of a = 1/10 and value of b = -1/10

So, Option A is correct.