For complex lovers
Let θ = ∣ m a x ( a r g ( R e ( z − i 1 ) . R e ( z ) + i I m ( z − i 1 ) ) ) ∣
where ∣ z ∣ = 1 and z = i and w = e i θ ,
if x = a w + b w 2 + c , y = a + b w + c w 2 , z = a w 2 + b + c w then
Find the value of ∣ a 2 + b 2 + c 2 x 2 + y 2 + z 2 + 2 a b + 2 b c + 2 a c ∣
(where a , b , c , x , y , z are non zero complex numbers).
The answer is 1.
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1 solution
I think the imaginary part would be -(y-1)/x^{2} But it doesn't affect the answer
Let z = x + i y
a r g ( R e ( z − i 1 ) . R e ( z ) + i I m ( z − i 1 )
= a r g ( R e ( x + i ( y − 1 ) 1 ) . x + i I m ( x + i ( y − 1 ) 1 )
= a r g ( R e ( x 2 + ( y − 1 ) 2 x − i ( y − 1 ) ) . x + i I m ( x 2 + ( y − 1 ) 2 x − i ( y − 1 ) )
= a r g ( x 2 + ( y − 1 ) 2 x . x − i x 2 + ( y − 1 ) 2 ( y − 1 ) )
= a r g ( x 2 + ( y − 1 ) 2 x 2 − x 2 + ( y − 1 ) 2 i ( y − 1 ) )
= a r c t a n ( x 2 − ( y − 1 ) )
m a x ( a r c t a n ( x 2 − ( y − 1 ) ) = 2 π
when x = 0 , y = − 1
⇒ w = i
Substituting the value of w in expression of x , y , z
we can easily get the answer as 1.