+185 Let x be a complex number such that the magnitude of x is 2. Find the largest possible distance between (3+4i)x^3 and x^5 when plotted in the complex plane.
I have referred to this link for help but not sure how to apply it to this problem. Help is appreciated https://artofproblemsolving.com/wiki/index.php/2012_AIME_II_Problems/Problem_6
Given that x is a complex number with magnitude 2, we know that |x| = 2. We can also express x in polar form as x = 2(cos θ + i sin θ) for some θ.
Now consider the expression (3 + 4i)x^3. We can write this as:
(3 + 4i)x^3 = (3 + 4i)2^3(cos 3θ + i sin 3θ).
Similarly, x^5 can be expressed as:
x^5 = 2^5(cos 5θ + i sin 5θ).
The distance between (3 + 4i)x^3 and x^5 is given by the magnitude of their difference, which is:
|(3 + 4i)x^3 - x^5| = |(3 + 4i)2^3(cos 3θ + i sin 3θ) - 2^5(cos 5θ + i sin 5θ)|
= |8(cos 3θ + i sin 3θ) - 32(cos 5θ + i sin 5θ) + 4i|
= |8 cos 3θ - 32 cos 5θ + 8i sin 3θ - 32i sin 5θ|
= 8|(cos 3θ - 4 cos 5θ) + i(sin 3θ - 4 sin 5θ)|.
To find the maximum value of this expression, we want to maximize the expression inside the absolute value, which can be thought of as the distance between two points in the complex plane
1
: (cos 3θ - 4 cos 5θ, sin 3θ - 4 sin 5θ) and (0,0). We need to find the angle θ that maximizes this distance.
This can be done by finding the angle between the two points, which is given by arctan((sin 3θ - 4 sin 5θ)/(cos 3θ - 4 cos 5θ)), and setting it equal to π. This is because the maximum distance between two points is achieved when the line connecting them is perpendicular to the x-axis, and hence has a slope of tan(π/2) = undefined.
Solving for θ yields:
θ = (1/2)(arctan(-4) + π/2)
Plugging this into our expression above gives an answer of 60.
