Mr. Giuseppe Cammarata at the Liceo Scientifico Statale Benedetto Croce in Palermo, Italy is using this material in his Course IV F. Many of the solutions below were written by his students.

1 (a) Let A and B be the projection of the point P to the real and the imaginary axes respectively. We see that is a 30o ,60o,90o triangle. The argument of the point P is and the hypotenuse has length 8. The segment is the imaginary part of the point P, while the segment is the real part of the point P. Then and Hence
(b) The point P is in the second quadrant and it has argument is a 30o, 60o, 90o triangle and the hypotenuse OP has length 8. The segment and the segment Then and Hence
(c) The point P is in the 3rd quadrant and it has argument We see that is a 45o, 45o, 90o triangle and the hypotenuse OP as length The two equal sides must have length Then and Hence P = A + B = -1 - i.
(d) The point P is on the imaginary axis and it has length 9. The real part has length 0, hence P = 0 - 9i = -9i.
(e) The point P is in the 4th quadrant, its argument is and the hypotenuse OP has length 4. We see that is a 45o, 45o, 90o triangle and the two sides have length Then and Hence

Solution by Francesco Felice and Daniele Manto
Liceo Scientifico Statale Benedetto Croce, May 2001.

9. To find the negative of a point in polar form we must add or subtract to the given angle. In rectangular form the negative of the complex number a + bi is the complex number -a - bi.

(a)

(b)

The conjugate of the point in polar form is the point that has the same absolute value as P but as argument the negative of that of P. In rectangular form the conjugate of the complex number a + bi is the complex number a - bi with the same real part and opposite imaginary part. So:

(c)

(d)

The product of two points and is the point that has as absolute value the product of the absolute values of P and Q, and as argument the sum of the arguments of P and Q. In rectangular form we find the product of two complex number using the following rule: (a + bi)(c + di) = (ac - bd) + (ad + bc)i.

(e)

(f)

Solution by Alessandro Belmonte and Mariano Brusca.
Liceo Scientifico Statale Benedetto Croce, April 2001.

17. (a) and iA = (0 + i)(3 - 4i) = 4 + 3i.

(b) and -A = -3 + 4i.

(d) and

(f) and A² = -7 - 24i.

(g) and A³ = -177 - 44i.

Solution by Rosalia Di Liberto and Rosa Maria Siggia
Liceo Scientifico Statale Benedetto Croce, May 2001.

21. a) because the conjugate of a point P is the point with the same real part as P but with imaginary part negative of that of P.

b) = (4 + 3i)(12 + 5i) = 48 + 20i + 36i - 15 = 33 + 56i.

d) Q² = (4 + 3i)² = 16 - 9 + 24i = 7 + 24i.

e) PQ = (12 - 5i)(4 + 3i) = 48 + 36i - 20i + 15 = 63 + 16i.

Solution by Gianluca Ridulfo and Sandro Caracausi
Liceo Scientifico Statale Benedetto Croce, April 2001.

23. and

Multiplying these gives us

Solution by Stefania Cuccia and Denada Rrushi
Liceo Scientifico Statale Benedetto Croce, May 2001.

25. To find a number B which at fifth power gives A, we have to take the fifth root of it, but there are five posible solutions. The operative formulas to take the 5th roots is: The solutions are:

We decided to choose the value of B for K = 0. Now we can make our multiplications:

Solution by Marco Modica and Fabrizio Martino
Liceo Scientifico Statale Benedetto Croce, May 2001.

31. This is the relation between c and s:

(1) s² = 1 - c².

It's the Theorem of Pythagoras.

We have two cases:

Case 1, if s > 0 the text indicates We square it to get In this equation we use relation (1) to get

Case 2, if s < 0 the text indicates We square it to get In this equation we use relation (1) and the fact that to get

Solution by Anna La Rocca and Marianna Bruno
Liceo Scientifico Statale Benedetto Croce, May 2001.