Question

Given that $\frac{1+i}{1+2^{2}i}\times \frac{1+3^{2}i}{1+4^{2}i}\times ....\times \frac{1+(2n-1{)}^{2}i}{1+(2n{)}^{2}i}=\frac{a+bi}{c+di},showthat:$
$\frac{2}{17}\times \frac{82}{257}\times ....\times \frac{(2n-1{)}^4+1}{(2n{)}^4+1}=\frac{a^2+b^2}{c^2+d^2}.$

Answer 1

$\frac{1+i}{1+2^{2}i}\times \frac{1+3^2}{1+4^{2}i}\times ...\times \frac{1+(2n-1{)}^{2}i}{1+(2n{)}^{2}i}=\frac{a+bi}{c+di}\Rightarrow \frac{1+i}{1+2^2}\times \frac{1+3^{2}i}{1+4^{2}i}\times .....\times \frac{1+(2n-1{)}^{2}i}{1+(2n{)}^{2}i}=\left|\frac{a+bi}{c+di}\right|\Rightarrow \left|\frac{1+i}{1+2^{2}i}\right|\times \left|\frac{1+3^{2}i}{1+4^{2}i}\right|\times .....\times \left|\frac{1+(2n-1{)}^{2}i}{1+(2n{)}^{2}i}\right|=\left|\frac{a+bi}{c+di}\right|.$
$\Rightarrow \frac{||1+i}{1+2^{2}i}.\frac{1+3^{2}i}{1+4^{2}i}\times ...\frac{|1+(2n-1{)}^{2}i|}{|1+(2n{)}^{2}i|}=\frac{|a+bi|}{|c+di|}\Rightarrow \frac{\sqrt{1+1}}{\sqrt{1+16}}.\frac{\sqrt{1+81}}{\sqrt{1+256}}....\frac{\sqrt{(2n-1{)}^4+1}}{\sqrt{(2n{)}^4+1}}=\frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}}\Rightarrow \frac{2}{17}.\frac{82}{257}...\frac{(2n-1{)}^4+1}{(2n{)}^4+1}=\frac{a^2+b^2}{c^2+d^2}.$