Answer

Verified

472.2k+ 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

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Mark and label the given geoinformation on the outline class 11 social science CBSE

When people say No pun intended what does that mea class 8 english CBSE

Name the states which share their boundary with Indias class 9 social science CBSE

Give an account of the Northern Plains of India class 9 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Trending doubts

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

At which age domestication of animals started A Neolithic class 11 social science CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Summary of the poem Where the Mind is Without Fear class 8 english CBSE

One cusec is equal to how many liters class 8 maths CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Change the following sentences into negative and interrogative class 10 english CBSE