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Sunny_sXe [5.5K]
3 years ago
10

Anybody know how to solve ??????

Mathematics
1 answer:
lys-0071 [83]3 years ago
3 0

Have you gone over sin cos and tan

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If (2i/2+i)-(3i/3+i)=a+bi, then a=<br>A. 1/10<br>B. -10<br>C. 1/50<br>D. -1/10​
Verizon [17]

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.

8 0
4 years ago
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