Question

If(1+i)(1+2i)(1+3i)... (1+ni)=(x+iy) then prove that 2x5x10x...×(1+n²) = (x² + y²).
koi answer de do. ​

Answer 1

$\large\underline{\sf{Solution-}}$

Given Complex number is

$\sf \: (1 + i)(1 + 2i)(1 + 3i) - - - (1 + ni) = (x + iy)$

Taking modulus on both sides, we get

$\sf \: \bigg |(1 + i)(1 + 2i)(1 + 3i) - - - (1 + ni)\bigg | = |x + iy|$

We know

$\boxed{ \tt{ \: |z_1 \times z_2| = |z_1| \times |z_2| \: }}$

and

$\boxed{ \tt{ \: z = x + iy \: \: \rm \implies\: |z| = \sqrt{ {x}^{2} + {y}^{2} } \: }}$

So, using these results, we get

$\sf \: |1 + i| |1 + 2i| |1 + 3i| - - - - |1 + ni| = \sqrt{ {x}^{2} + {y}^{2} }$

can be further rewritten as

$\sf \:(\sqrt{ {1}^{2}+{1}^{2} })( \sqrt{ {1}^{2}+{2}^{2} })( \sqrt{ {1}^{2} +{3}^{2} }) - - ( \sqrt{1+{n}^{2}}) = \sqrt{ {x}^{2} + {y}^{2} }$

can be further rewritten as

$\sf \: \sqrt{2} \times \sqrt{5} \times \sqrt{10} \times - - - \times \sqrt{1 + {n}^{2} } = \sqrt{ {x}^{2} + {y}^{2} }$

On squaring both sides, we get

$\boxed{ \tt{ \: 2 \times 5 \times 10 \times - - - \times ( {n}^{2} + 1) = {x}^{2} + {y}^{2} \: }}$

Hence, Proved

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Additional Information :-

$\red{\rm :\longmapsto\:z \: \overline{z} = { |z| }^{2} \: }$

$\red{\rm :\longmapsto\: |z| = |\overline{z}| \: }$

$\red{\rm :\longmapsto\:\overline{z_1 + z_2} = \overline{z_1} + \overline{z_2} \: }$

$\red{\rm :\longmapsto\:\overline{z_1 - z_2} = \overline{z_1} - \overline{z_2} \: }$

$\red{\rm :\longmapsto\:\overline{z_1 \times z_2} = \overline{z_1} \times \overline{z_2} \: }$

$\red{\rm :\longmapsto\:\overline{z_1 \div z_2} = \overline{z_1} \div \overline{z_2} \: }$