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Solutions to Math Complex Problems

1. (D)

2. (B) Two given points, 1 + 2i and -1 - 2i, are symmetric with respect to the origin,
and the distances from the origin to all three points are equal. Therefore the origin is the
center of the square, and the fourth vertex must be symmetric to -2 + i around it. Thus
the fourth vertex is 2 - i.

3. (B) We compute ,and .Since , it
follows that , which is units from the origin.

4. (C) Write F as a + bi, where we see from the diagram that a, b > 0 and a2 + b2 > 1.

we see that the reciprocal of F is in quadrant IV, since the real part on the right is positive
and the coefficient on the imaginary part is negative. Also, the magnitude of the reciprocal

Thus the only possibility is point C.

5. (D) The set S consists of all complex numbers of the form

for some real number r. Since S consists of all real multiples of 3 - 4i, each point in S is
on the line through the origin and 3 - 4i, and conversely.

6. (B) Use the quadratic formula to obtain

which has discriminant D = -9 + 40k. If k = 1, then D = 31, so (A) is false. If k is
a negative real number, then D is a negative real number, so (B) is true. If k = i, then
D = -9+40i = 16+40i-25 = (4+5i)2, and the roots are 1/5+(2/5)i and -1/5-(1/10)i,
so (C) and (D) are false. If k = 0 (which is a complex number), then the roots are 0 and
(3/10)i, so (E) is false.

7. (D) Recall, from DeMoivre's Theorem, that the six sixth roots of -64 are equispaced
around the circle of radius . Since ±2i are roots, exactly two of the roots are in the
right half-plane and they must be conjugates. The product of any pair of conjugates is the
square of their distance from the origin, so the product of these two roots is .

8. (E) Set z = a + bi. We seek .So

Thus, a2 + b2 = 225 + 64 = 289.

9. (B) The quadratic formula leads to the roots

To find , let (a + bi)2 = 21 - 20i where a and b are real. Equating real and
imaginary parts leads to a2-b2 = 21 and 2ab = -20. Solve these equations simultaneously:

so a = ±5 and b = 2. Thus a + bi = 5- 2i or -5 + 2i. Therefore,

The product of the real parts of these two roots is -6.

10. (C) Note that x3 = y3 = 1, because x and y are the complex roots of 0 = x3- 1 =
(x-1)(x2+x+1). Alternatively, plot x and y in the complex plane and observe that they
have modulus 1 and arguments 2π /3 and 4π /3, Thus x9 + y9 = 1+1 ≠ -1.