A) \[-1\]
B) 1
C) \[1/2\]
D) \[-1/2\]
Correct Answer: B
Solution :
\[z=\frac{(1+i)(2+i)}{3+i}\times \frac{3-i}{3-i}\] \[=\frac{(2+i+2i+{{i}^{2}})(3-i)}{9-{{i}^{2}}}\] \[=\frac{(1+3i)(3-i)}{9+1}\] \[=\frac{3-i+9i-3{{i}^{2}}}{10}\] \[z=\frac{6+8i}{10}=\frac{3}{5}+\frac{4}{5}i\] \[|z|=\sqrt{{{\left( \frac{3}{2} \right)}^{2}}+{{\left( \frac{4}{5} \right)}^{2}}}\] \[=\sqrt{\frac{9}{25}+\frac{16}{25}}=\sqrt{\frac{25}{25}}=1\]
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