Problem:
A region S in the complex plane is defined by
S={x+iy:−1≤x≤1,−1≤y≤1}
A complex number z=x+iy is chosen uniformly at random from S. What is the probability that (43+43i)z is also in S ?
Answer Choices:
A. 21
B. 32
C. 43
D. 97
E. 87 Solution:
Let f(z)=(43+43i)z. The effect of multiplying z by (43+43i) is to rotate z an angle equal to arg(43+43i)=4π from the origin, and to magnify by a factor of ∣∣∣∣∣43+43i∣∣∣∣∣=432. Thus the image S′ of S under f is a square region with vertices ±23 and ±23i. The area of S′ is (432⋅2)2. The intersection of S and S′ is an octagonal region obtained from S′ by removing four congruent triangular regions. The topmost of these triangles T has vertices 21+i,23i, and −21+i, so its area equals 41. Then the requested probability is
(432⋅2)2(432⋅2)2−4⋅41=97
\section*{OR}
The product is (43+43i)(x+iy)=(43x−43y)+(43x+43y)i. The point x+iy will be in S if and only if −1≤43x−43y≤1 and −1≤43x+43y≤1, which are equivalent to −34≤x−y≤34 and −34≤x+y≤34. Thus x+yi must be inside the square with vertices ±34 and ±34i. By symmetry we can look at just the first quadrant. Because the portion of S in the first quadrant has area 1, the desired probability is the area of the portion of the interior of this square within S. The squares intersect at 1+31i and 31+i, so the desired probability is 1−21⋅32⋅32=97.
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