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

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

Answer

Verified

360.3k+ views

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

Last updated date: 24th Sep 2023

•

Total views: 360.3k

•

Views today: 5.60k

Recently Updated Pages

What do you mean by public facilities

Difference between hardware and software

Disadvantages of Advertising

10 Advantages and Disadvantages of Plastic

What do you mean by Endemic Species

What is the Botanical Name of Dog , Cat , Turmeric , Mushroom , Palm

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Drive an expression for the electric field due to an class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

What is the past tense of read class 10 english CBSE