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
•
Total views: 303k
•
Views today: 2.83k
Answer
303k+ 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
Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE
