MCQ
If $\text{x}+\text{iy}=(1+\text{i})(1+2\text{i})(1+3\text{i}),$ then $\text{x}^2+\text{y}^2=$
- A0
- B1
- ✓100
- Dnone of these
✓
Answer
Correct option: C.
100
$\therefore\text{x}+\text{iy}=(1+\text{i})(1+2\text{i})(1+3\text{i})$
Taking modulus on both the sides:
$|\text{x}+\text{iy}|=|(1+\text{i})(1+2\text{i})(1+3\text{i})|$
$\Rightarrow|\text{x}+\text{iy}|=|(1+\text{i})|\times|(1+2\text{i})|\times|(1+3\text{i})|$
$\Rightarrow\sqrt{\text{x}^2+\text{y}^2}=\sqrt{1^2+1^2}\sqrt{1^2+2^2}\sqrt{1^2+3^2}$
$\Rightarrow\sqrt{\text{x}^2+\text{y}^2}=\sqrt{2}\sqrt{5}\sqrt{10}$
$\Rightarrow\sqrt{\text{x}^2+\text{y}^2}=\sqrt{100}$
Squaring both the sides,
$\Rightarrow\text{x}^2+\text{y}^2=100$
Taking modulus on both the sides:
$|\text{x}+\text{iy}|=|(1+\text{i})(1+2\text{i})(1+3\text{i})|$
$\Rightarrow|\text{x}+\text{iy}|=|(1+\text{i})|\times|(1+2\text{i})|\times|(1+3\text{i})|$
$\Rightarrow\sqrt{\text{x}^2+\text{y}^2}=\sqrt{1^2+1^2}\sqrt{1^2+2^2}\sqrt{1^2+3^2}$
$\Rightarrow\sqrt{\text{x}^2+\text{y}^2}=\sqrt{2}\sqrt{5}\sqrt{10}$
$\Rightarrow\sqrt{\text{x}^2+\text{y}^2}=\sqrt{100}$
Squaring both the sides,
$\Rightarrow\text{x}^2+\text{y}^2=100$
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