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Home  >>  TN XII Math  >>  Complex Numbers
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If $\left ( 1+\mathit{i} \right )\left ( 1+2\mathit{i} \right )\left ( 1+3\mathit{i} \right )... \left ( 1+\mathit{ni} \right )=\mathit{x+iy}$, show that 2.5.10 ... $\left ( 1+n^{2} \right )$ = $ x^{2}+y^{2}$

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1 Answer

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Toolbox:
  • If $z=a+ib$ then ,
  • $\bar{z}=a-ib$
  • $\mid z\mid=\sqrt{a^2+b^2}$
  • $z^{-1}=\large\frac{a-ib}{a^2+b^2}$
  • $z\bar{z}=a^2+b^2$
  • Also $Re(z)=a,Im(z)=b$
  • If $z_1=a+ib,z_2=c+id$
  • $z_1z_2=(a+ib)(c+id)=(ac-bd)+i(ad+bc)$
  • $\mid z_1z_2\mid=\mid z_1\mid\mid z_2\mid$
Step 1:
$(1+i)(1+2i)(1+3i)......(1+ni)=(x+iy)$
Equating the moduli,
$\mid (1+i)(1+2i)(1+3i).....(1+ni)\mid=\mid x+iy\mid $
$\mid 1+i\mid\mid 1+2i\mid\mid 1+3i\mid.....\mid 1+ni \mid=\mid x+iy\mid $
Step 2:
Squaring we get
$\mid 1+i\mid^2\mid 1+2i\mid^2\mid 1+3i\mid^2.....\mid 1+ni \mid^2=\mid x+iy\mid ^2$
$(1+1)(1+4)(1+9).....(1+n^2)=x^2+y^2$
$2.5.10..........(1+n^2)=x^2+y^2$
answered Jun 7, 2013 by sreemathi.v
 

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