can someone help me find a complex number w/ an absolute value between 3 and 8 please? im confused!
2007-12-16 01:50:34 · 5 answers · asked by anna 1 in Science & Mathematics ➔ Mathematics
complex number = x +iy
absolute value of a complex number is
sqrt (x^2 +y^2)
you want
3 < sqrt(x^2 +y^2) <8
square all terms to get
9 < x^2 +y^2 <64
think of these as 2 concentric circles with origin at (0,0) and radii = 3 and 8
if you draw the 2 circles
circle1: radius = 3 and center (0.0)
circle2: radius =8 and center (0,0)
any point between the 2 circles has an absolute value between 3 and 8.
2007-12-16 02:30:53 · answer #1 · answered by Anonymous · 0⤊ 0⤋
4 has absolute value 4
also 5, 6 and 7 have absolute values 5, 6 and 7 respectively
2007-12-16 03:27:56 · answer #2 · answered by Theta40 7 · 0⤊ 0⤋
The absolute value of a + bi is √(a+bi)(a-bi) = √(a²+ b²)
So look for a,b,c such that a²+b² = c².
a+ bi will be the number you want and c will be its
absolute value.
So 3+4i has absolute value 5, which satisfies
your requirements.
2007-12-16 02:33:24 · answer #3 · answered by steiner1745 7 · 0⤊ 0⤋
How about 3+4i
absolute value of this is 5
(by pythagoras theorem)
2007-12-16 01:57:30 · answer #4 · answered by Anonymous · 0⤊ 0⤋
there are a lot of choices as long as 3< a^2 + b^2 < 8 1+2i, 2 + i etc.
2016-05-24 04:53:10 · answer #5 · answered by ? 3 · 0⤊ 0⤋