Complex numbers with integers

Algebra Level 5

If a , b , c a,b,c are distinct integers and ω \omega is cube root of unity, then find the minimum value of a + b ω + c ω {|a+b\omega +\frac{c}{\omega}|} . Give your answer to 3 decimal places.


The answer is 1.732.

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1 solution

Tanishq Varshney
May 13, 2015

l e t ω b e w let~\omega~be~w ,just to avoid typing.

The expression is a + b w + c w 2 |a+bw+cw^{2}| as 1 w = w 2 \frac{1}{w}=w^{2}

a + b w + c w 2 2 = ( a + b w + c w 2 ) ( a + b w + c w 2 ) |a+bw+c{ w }^{ 2 }|^{ 2 }=(a+bw+c{ w }^{ 2 })(\overline { a+bw+c{ w }^{ 2 } } )

a + b w + c w 2 2 = a 2 + b 2 + c 2 ( a b + b c + c a ) |a+bw+c{ w }^{ 2 }|^{ 2 }=a^{2}+b^{2}+c^{2}-(ab+bc+ca)

using w 3 = 1 a n d w + w 2 = 1 w^{3}=1~and~w+w^{2}=-1 we get the above expression

a + b w + c w 2 2 = 1 2 ( ( a b ) 2 + ( b c ) 2 + ( c a ) 2 ) \large{|a+bw+c{ w }^{ 2 }|^{ 2 }=\frac{1}{2}((a-b)^{2}+(b-c)^{2}+(c-a)^{2})}

since a , b , c a,b,c are distinct integers m i n ( a b ) = 1 a n d m i n ( b c ) = 1 a n d m i n ( c a ) = 2 min(a-b)=1~and~min(b-c)=1~and~min(c-a)=2

or u can put a = 1 , b = 2 , c = 3 a=1,~b=2~,c=3

m i n a + b w + c w 2 2 = 1 2 ( 1 + 1 + 4 ) \large{min|a+bw+c{ w }^{ 2 }|^{ 2 }=\frac{1}{2}(1+1+4)}

m i n a + b w + c w 2 = 3 \huge{min|a+bw+c{ w }^{ 2 }|=\sqrt{3}}

cheers!!

Nice problem! Did exactly what the solution says. Thoroughly enjoyed it. :D

Sanchit Aggarwal - 5 years, 7 months ago

Nicely done. Usually many take c-a=1 just like i did.☺

Kyle Finch - 6 years, 1 month ago

Nice work , A similar prob. was asked in JEE. But it said "Not all same". Good observation !

Keshav Tiwari - 6 years, 1 month ago

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yeah,had just done that jee problem how was your advanced?

NILAY PANDE - 6 years ago

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Let's wait for the results .

Keshav Tiwari - 6 years ago

@Tanishq Varshney Nice problem nice solution! +!

User 123 - 5 years, 12 months ago

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