Answer:
Option A is correct.
Explanation:
We are given:
![(2i)/(2+i)-(3i)/(3+i) = a+bi](https://img.qammunity.org/2020/formulas/mathematics/middle-school/cpgb9w3crljwvmussmwkk03oq3cpvpguvy.png)
We need to find the value of a.
The LCM of (2+i) and (3+i) is (2+i)(3+i)
![=(2i(3+i))/((2+i)(3+i))-(3i(2+i))/((2+i)(3+i))\\=(6i+2i^2)/((2+i)(3+i))-(6i+3i^2)/((2+i)(3+i))\\=(6i+2i^2-(6i+3i^2))/((2+i)(3+i))\\=(6i+2i^2-6i-3i^2))/(5+5i)\\=(-i^2)/(5+5i)\\i^2=-1\\=(-(-1))/(5+5i)\\=(1)/(5+5i)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/mrze7f4eod4xggfpziz1xu9zjmu8vdbaqu.png)
Now rationalize the denominator by multiplying by 5-5i/5-5i
![=(1)/(5+5i)*(5-5i)/(5-5i) \\=(5-5i)/((5+5i)(5-5i))\\=(5-5i)/((5+5i)(5-5i))\\(a+b)(a-b)= a^2-b^2\\=(5(1-i))/((5)^2-(5i)^2)\\=(5(1-i))/(25+25)\\=(5(1-i))/(50)\\=(1-i)/(10)\\=(1)/(10)-(i)/(10)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/o44qye1fuyxja3czo0om0wg2zf92hs3eda.png)
We are given
![(2i)/(2+i)-(3i)/(3+i) = a+bi](https://img.qammunity.org/2020/formulas/mathematics/middle-school/cpgb9w3crljwvmussmwkk03oq3cpvpguvy.png)
Now after solving we have:
![(1)/(10)-(i)/(10)=a+bi](https://img.qammunity.org/2020/formulas/mathematics/middle-school/jbmao4mqijfmvhxfewt20yhhq1vl7mq9ov.png)
So value of a = 1/10 and value of b = -1/10
So, Option A is correct.