Question

If $x+iy=(1+i)(1+2i)(1+3i)$, then x² + y² =
(a) 0
(b) 1
(c) 100
(d) none of these

Answer 1

(c) 100

$$\begin{gathered} \because x+iy=(1+i)(1+2i)(1+3i) \\ \mathrm{Taking}\mathrm{modulus}\mathrm{on}\mathrm{both}\mathrm{the}\mathrm{sides}: \\ \left|x+iy\right|=\left|(1+i)(1+2i)(1+3i)\right| \\ \Rightarrow \left|x+iy\right|=\left|1+i\right|\times \left|1+2i\right|\times \left|1+3i\right| \\ \Rightarrow \sqrt{x^2+y^2}=\sqrt{1^2+1^2}\sqrt{1^2+2^2}\sqrt{1^2+3^2} \\ \Rightarrow \sqrt{x^2+y^2}=\sqrt{2}\sqrt{5}\sqrt{10} \\ \Rightarrow \sqrt{x^2+y^2}=\sqrt{100} \\ \mathrm{Squaring}\mathrm{both}\mathrm{the}\mathrm{sides}, \\ \Rightarrow x^2+y^2=100 \end{gathered}$$