MHT CET · Maths · Complex Number
If \(\frac{3+2 \mathrm{i}}{1+\mathrm{i}}=\frac{1}{2}(\mathrm{x}+\mathrm{iy})\), then \(\mathrm{x}-\mathrm{y}=\)
- A 4
- B 3
- C 6
- D 5
Answer & Solution
Correct Answer
(C) 6
Step-by-step Solution
Detailed explanation
\(\frac{3+2 \mathrm{i}}{1+\mathrm{i}}=\frac{1}{2}(\mathrm{x}+\mathrm{iy})\)
\(\therefore \mathrm{x}+\mathrm{iy}=\frac{2(3+2 \mathrm{i})}{1+\mathrm{i}} \times \frac{1-\mathrm{i}}{1-\mathrm{i}}=\frac{(6+4 \mathrm{i})(1-\mathrm{i})}{1-\mathrm{i}^2}\)
\(=\frac{6+4 \mathrm{i}-6 \mathrm{i}-4 \mathrm{i}^2}{1+1}=5-\mathrm{i}\)
\(\therefore \mathrm{x}=5, \mathrm{y}=-1 \Rightarrow \mathrm{x}-\mathrm{y}=6\)
\(\therefore \mathrm{x}+\mathrm{iy}=\frac{2(3+2 \mathrm{i})}{1+\mathrm{i}} \times \frac{1-\mathrm{i}}{1-\mathrm{i}}=\frac{(6+4 \mathrm{i})(1-\mathrm{i})}{1-\mathrm{i}^2}\)
\(=\frac{6+4 \mathrm{i}-6 \mathrm{i}-4 \mathrm{i}^2}{1+1}=5-\mathrm{i}\)
\(\therefore \mathrm{x}=5, \mathrm{y}=-1 \Rightarrow \mathrm{x}-\mathrm{y}=6\)
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