# If (1+i)(1+2i)…(1+ni) = x+iy, then prove that

\[{{x}^{2}}+{{y}^{2}}=2\cdot 5\cdot 10\ldots \left( 1+{{n}^{2}} \right)\]

Last updated date: 14th Mar 2023

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Answer

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Hint: Apply mod on both sides of the equation. Use the fact that if a and b are two complex numbers then |ab| = |a||b|. Use the fact that if z = x+iy then $\left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}$. Square both sides and use the fact that \[{{\left( ab \right)}^{m}}={{a}^{m}}{{b}^{m}}\]

Complete step-by-step answer:

We have (1+i)(1+2i)…(1+ni) = x+iy,

Taking absolute value on both sides, we get

|(1+i)(1+2i)…(1+ni)| =| x+iy|

Using |ab| = |a||b|, we get

|(1+i)||(1+2i)|…|(1+ni)| = |x+iy|

Squaring both sides, we get

\[{{\left( \left| \left( 1+i \right) \right|\left| \left( 1+2i \right) \right|\ldots \left| \left( 1+ni \right) \right| \right)}^{2}}\text{ }=\text{ }{{\left( \left| x+iy \right| \right)}^{2}}\]

Using if z = x+iy then $\left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}$, we get

\[\begin{align}

& {{\left( \sqrt{{{1}^{2}}+{{1}^{2}}}\sqrt{{{1}^{2}}+{{2}^{2}}}\ldots \sqrt{{{1}^{2}}+{{n}^{2}}} \right)}^{2}}\text{ }=\text{ }{{\left( \sqrt{{{x}^{2}}+{{y}^{2}}} \right)}^{2}} \\

& \Rightarrow \left( 1+1 \right)\left( 1+4 \right)\ldots \left( {{1}^{2}}+{{n}^{2}} \right)={{x}^{2}}+{{y}^{2}} \\

& \Rightarrow 2\cdot 5\ldots \left( {{1}^{2}}+{{n}^{2}} \right)={{x}^{2}}+{{y}^{2}} \\

\end{align}\]

Hence proved.

Note: [1] The above statement can also be proved by taking the conjugate over the whole expression and using $z\overline{z}={{\left| z \right|}^{2}}$

We have (1+i)(1+2i)…(1+ni) = x+iy (i)

Taking conjugate on both sides we get

\[\overline{~\left( 1+i \right)}\overline{\left( 1+2i \right)}\ldots \overline{\left( 1+ni \right)}\text{ }=\text{ }\overline{x+iy}\text{ (ii)}\]

Multiplying equation (i) and equation (ii) we get

\[\begin{align}

& \overline{~\left( 1+i \right)}\overline{\left( 1+2i \right)}\ldots \overline{\left( 1+ni \right)}(1+i)\left( 1+2i \right)\ldots \left( 1+ni \right)\text{ = }\overline{x+iy}\left( x+iy \right) \\

& \Rightarrow \left( 1+i \right)\left( 1+i \right)\overline{\left( 1+2i \right)}\left( 1+2i \right)\ldots \overline{\left( 1+ni \right)}\left( 1+ni \right)={{x}^{2}}+{{y}^{2}} \\

& \Rightarrow \left( 1+1 \right)\left( 1+4 \right)\ldots \left( {{1}^{2}}+{{n}^{2}} \right)={{x}^{2}}+{{y}^{2}} \\

& \Rightarrow 2\cdot 5\ldots \left( {{1}^{2}}+{{n}^{2}} \right)={{x}^{2}}+{{y}^{2}} \\

\end{align}\]

Hence proved.

[2] The property |ab|= |a||b| of complex numbers are very important and have a lot of applications in the field of mathematics,e.g. It can be proven that the product of numbers expressible as the sum of two squares is also expressible as the sum of two squares. Consider number P and Q such that $P={{a}^{2}}+{{b}^{2}}$ and $Q={{c}^{2}}+{{d}^{2}}$

Let ${{z}_{1}}=a+ib$ and ${{z}_{2}}=c+id$

${{z}_{1}}{{z}_{2}}=ac-bd+i\left( ad+bc \right)$

Taking mod on both sides we get

$\left| {{z}_{1}} \right|\left| {{z}_{2}} \right|=\sqrt{{{\left( ac-bd \right)}^{2}}+{{\left( ad+bc \right)}^{2}}}$

Squaring both sides we get

$PQ={{\left( ac-bd \right)}^{2}}+{{\left( ad+bc \right)}^{2}}$

In other words, PQ is also a sum of squares of two numbers

Complete step-by-step answer:

We have (1+i)(1+2i)…(1+ni) = x+iy,

Taking absolute value on both sides, we get

|(1+i)(1+2i)…(1+ni)| =| x+iy|

Using |ab| = |a||b|, we get

|(1+i)||(1+2i)|…|(1+ni)| = |x+iy|

Squaring both sides, we get

\[{{\left( \left| \left( 1+i \right) \right|\left| \left( 1+2i \right) \right|\ldots \left| \left( 1+ni \right) \right| \right)}^{2}}\text{ }=\text{ }{{\left( \left| x+iy \right| \right)}^{2}}\]

Using if z = x+iy then $\left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}$, we get

\[\begin{align}

& {{\left( \sqrt{{{1}^{2}}+{{1}^{2}}}\sqrt{{{1}^{2}}+{{2}^{2}}}\ldots \sqrt{{{1}^{2}}+{{n}^{2}}} \right)}^{2}}\text{ }=\text{ }{{\left( \sqrt{{{x}^{2}}+{{y}^{2}}} \right)}^{2}} \\

& \Rightarrow \left( 1+1 \right)\left( 1+4 \right)\ldots \left( {{1}^{2}}+{{n}^{2}} \right)={{x}^{2}}+{{y}^{2}} \\

& \Rightarrow 2\cdot 5\ldots \left( {{1}^{2}}+{{n}^{2}} \right)={{x}^{2}}+{{y}^{2}} \\

\end{align}\]

Hence proved.

Note: [1] The above statement can also be proved by taking the conjugate over the whole expression and using $z\overline{z}={{\left| z \right|}^{2}}$

We have (1+i)(1+2i)…(1+ni) = x+iy (i)

Taking conjugate on both sides we get

\[\overline{~\left( 1+i \right)}\overline{\left( 1+2i \right)}\ldots \overline{\left( 1+ni \right)}\text{ }=\text{ }\overline{x+iy}\text{ (ii)}\]

Multiplying equation (i) and equation (ii) we get

\[\begin{align}

& \overline{~\left( 1+i \right)}\overline{\left( 1+2i \right)}\ldots \overline{\left( 1+ni \right)}(1+i)\left( 1+2i \right)\ldots \left( 1+ni \right)\text{ = }\overline{x+iy}\left( x+iy \right) \\

& \Rightarrow \left( 1+i \right)\left( 1+i \right)\overline{\left( 1+2i \right)}\left( 1+2i \right)\ldots \overline{\left( 1+ni \right)}\left( 1+ni \right)={{x}^{2}}+{{y}^{2}} \\

& \Rightarrow \left( 1+1 \right)\left( 1+4 \right)\ldots \left( {{1}^{2}}+{{n}^{2}} \right)={{x}^{2}}+{{y}^{2}} \\

& \Rightarrow 2\cdot 5\ldots \left( {{1}^{2}}+{{n}^{2}} \right)={{x}^{2}}+{{y}^{2}} \\

\end{align}\]

Hence proved.

[2] The property |ab|= |a||b| of complex numbers are very important and have a lot of applications in the field of mathematics,e.g. It can be proven that the product of numbers expressible as the sum of two squares is also expressible as the sum of two squares. Consider number P and Q such that $P={{a}^{2}}+{{b}^{2}}$ and $Q={{c}^{2}}+{{d}^{2}}$

Let ${{z}_{1}}=a+ib$ and ${{z}_{2}}=c+id$

${{z}_{1}}{{z}_{2}}=ac-bd+i\left( ad+bc \right)$

Taking mod on both sides we get

$\left| {{z}_{1}} \right|\left| {{z}_{2}} \right|=\sqrt{{{\left( ac-bd \right)}^{2}}+{{\left( ad+bc \right)}^{2}}}$

Squaring both sides we get

$PQ={{\left( ac-bd \right)}^{2}}+{{\left( ad+bc \right)}^{2}}$

In other words, PQ is also a sum of squares of two numbers

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