Problem:
A region S in the complex plane is defined by
S={x+iy:β1β€xβ€1,β1β€yβ€1}
A complex number z=x+iy is chosen uniformly at random from S. What is the probability that (43β+43βi)z is also in S ?
Answer Choices:
A. 21β
B. 32β
C. 43β
D. 97β
E. 87β Solution:
Let f(z)=(43β+43βi)z. The effect of multiplying z by (43β+43βi) is to rotate z an angle equal to arg(43β+43βi)=4Οβ from the origin, and to magnify by a factor of β£β£β£β£β£β43β+43βiβ£β£β£β£β£β=43β2β. Thus the image Sβ² of S under f is a square region with vertices Β±23β and Β±23βi. The area of Sβ² is (43β2ββ 2)2. The intersection of S and Sβ² is an octagonal region obtained from Sβ² by removing four congruent triangular regions. The topmost of these triangles T has vertices 21β+i,23βi, and β21β+i, so its area equals 41β. Then the requested probability is
\section*{OR}
The product is (43β+43βi)(x+iy)=(43βxβ43βy)+(43βx+43βy)i. The point x+iy will be in S if and only if β1β€43βxβ43βyβ€1 and β1β€43βx+43βyβ€1, which are equivalent to β34ββ€xβyβ€34β and β34ββ€x+yβ€34β. Thus x+yi must be inside the square with vertices Β±34β and Β±34βi. 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+31βi and 31β+i, so the desired probability is 1β21ββ 32ββ 32β=97ββ.
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