If x+iy=(1+i)(1+2i)(1+3i),
Question:

If $x+i y=(1+i)(1+2 i)(1+3 i)$, then $x^{2}+y^{2}=$

(a) 0

(b) 1

(c) 100

(d) none of these

Solution:

(c) 100

$\because x+i y=(1+i)(1+2 i)(1+3 i)$

Taking modulus on both the sides:

$|x+i y|=|(1+i)(1+2 i)(1+3 i)|$

$\Rightarrow|x+i y|=|1+i| \times|1+2 i| \times|1+3 i|$

$\Rightarrow \sqrt{x^{2}+y^{2}}=\sqrt{1^{2}+1^{2}} \sqrt{1^{2}+2^{2}} \sqrt{1^{2}+3^{2}}$

$\Rightarrow \sqrt{x^{2}+y^{2}}=\sqrt{2} \sqrt{5} \sqrt{10}$

$\Rightarrow \sqrt{x^{2}+y^{2}}=\sqrt{100}$

Squaring both the sides,

$\Rightarrow x^{2}+y^{2}=100$