MATH198 Solution Sheet 2
1. Use the quadratic formula to solve the equations √ (i) x2 + 12x + 27 = 0 ⇒ x = 12 −12 ± 36 = −6 ± 3 √ (ii) x2 + 12x + 40 = 0 ⇒ x = 21 −12 ± −16 = −6 ± 2i √ (iii) x2 + 2ix + 3 = 0 ⇒ x = 12 −2i ± −16 = −i ± 2i √ √ (iv) x2 − 2ix + 6 = 0 ⇒ x = 21 2i ± −28 = i ± i 7 2. Express the following complex numbers in the polar form reiθ √ This problem is done using, if z = x + iy, then r = x2 + y 2 and tan(θ) = y/x. However, there are always two choices for θ, both giving the correct tan. We have to be careful to choose the angle in the correct quadrant (see figure). Im 2i
(ii)
(i) i
-2
-1
1
-i
2
Re
(iv)
-2i
-3i (iii)
√ √ √ (i) z = 1 + i 3. r = 1 + 3 = 2 and θ = tan−1 ( 3/1) = π3 + kπ, (for any integer k). Check the quadrant: Both the real and imaginary parts are positive, so the angle must lie between 0 and π/2, so the correct choice is θ = π3 . √ π 1 + i 3 = 2ei 3 √ √ 2 ) = − π4 + kπ. This time (ii) z = −2 + 2i. then r = 22 + 22 = 2 2 and θ = tan−1 ( (−2) the real part is negative, so the calculator’s answer to tan−1 (−1) = −π/4 is the wrong
choice, we need − π4 + π = 43 π. −2 + 2i =
√
√ 3 3 8ei 4 π = 2 2ei 4 π
(see the figure). (iii) For z = −3i, it is easier to draw a diagram then to use the formulae to get the integral. On an Argand diagram −3i lies along the negative imaginary axis. Hence the angle is − π2 (or + 3π ). 2 π −3i = 3e−i 2 √ √ √ (iv) For z = 3 − i, then r = 1 + 3 = 2 and θ = tan−1 (−1/ 3) = − π6 + kπ. The real part of z is positive, so we want an angle in the range −π/2 to π/2, so this time the calculator’s answer − π6 is the one we want. √
π
3 − i = 2e−i 6
Lots of people got the angles wrong. Draw a picture of the complex plane, plot the point, and decide which quarter your point is in. 3. Show that sin 3θ = 3 sin θ − 4 sin3 θ . To solve this problem we use Euler’s remarkable formula raised to the third power. exp (3iθ) = cos(3θ) + i sin(3θ) exp (3iθ) = (exp (iθ))3 = (cos θ + i sin θ)3
(1) (2)
The strategy is to equate real and imaginary parts in equations (1) and (2). (de Moivre) (cos θ + i sin θ)3 = cos3 θ + 3 cos2 θ(i sin θ) + 3 cos θ(i sin θ)2 + (i sin θ)3 = cos3 θ + 3i cos2 θ sin θ − 3 cos θ sin2 θ − i sin3 θ Im[cos 3θ + i sin 3θ] = Im[(cos θ + i sin θ)3 ] ⇒ sin 3θ = 3 cos2 θ sin θ − sin3 θ = 3(1 − sin2 θ) sin θ − sin3 θ ⇒
sin 3θ =
3 sin θ − 4 sin3 θ.
4. Show that cos5 θ =
1 (cos 5θ + 5 cos 3θ + 10 cos θ) . 16
Use the exponential definition of cos θ cos θ =
1 iθ e + e−iθ 2
⇒
cos mθ =
1 imθ e + e−imθ 2
Now expand out cos5 θ using (a + b)5 = a5 + 5a4 b1 + 10a3 b2 + 10a2 b3 + 5a1 b4 + b5 (from Pascal’s triangle). cos5 θ = 1 25 1 = 5 2 1 = 4 2
=
⇒
1 iθ −iθ 5 e + e 25
5
eiθ + 5 eiθ
4
1
e−iθ + 10 eiθ
3
2
e5iθ + 5e3iθ + 10eiθ + 10e−iθ + 5e−3iθ + e−5iθ
e−iθ + 10 eiθ
2
1 5iθ 1 −5iθ 5 3iθ 5 −3iθ 10 iθ 10 −iθ e + e + e + e + e + e 2 2 2 2 2 2 1 (cos 5θ + 5 cos 3θ + 10 cos θ) cos5 θ = 16
3
e−iθ + 5 eiθ
1
4
e−iθ + e−iθ
5