MATH198 Solution Sheet 1
1. Given that z1 = 3 − 2i and z2 = 2 + i (a) determine (i) |z1 | , (ii) Re z1 , (iii) Im z2 . (i) |z1 | =
q
32 + (−2)2 =
√
13 .
Note: Not
q
32 + (−2i)2 .
(ii) Re z1 = 3 (iii) Im z2 = 1 .
By definition, Im(a + ib) ≡ b.
(b) Determine in the form a + ib 1 (i) 2z1 + 3z2 , (ii) z1 z2 , (iii) , z1
(iv)
z1 . z2
(i) 2z1 + 3z2 = 2(3 − 2i) + 3(2 + i) = 6 − 4i + 6 + 3i = 12 − i (ii) z1 z2 = (3 − 2i)(2 + i) = 6 + 3i − 4i − 2i2 = 8 − i (iii) (iv)
1 1 3 + 2i 3 + 2i 3 2 1 = = = = + i z1 3 − 2i 3 − 2i 3 + 2i 9+4 13 13
3 − 2i (3 − 2i)(2 − i) 4 − 7i 4 7 z1 = = = = − i z2 2+i (2 + i)(2 − i) 5 5 5
(c) Find the complex conjugate of z1 − z2 .
Taking the complex conjugate means reversing the sign of the imaginary part: z1 − z2 = (3 − 2i) − (2 + i) = 1 − 3i ⇒ (z1 − z2 )⋆ = 1 + 3i 2. Simplify (i) i6 , (ii) (1 + i)(1 − i),
(iii) i−5 .
(i) i6 = i4 i2 = 1 × (−1) = −1 (ii) (1 + i)(1 − i) = 1 + i − i − i2 = 2 (iii) i−5 = i−4 i−1 =
1 = −i i
−i is a “better” answer than 1i , (because −i fits the standard form a + ib).
3. Express the complex number (1 + i)(2 − 2i) +1 (3 + i)(2 − i)
in the form a + ib.
4 4(7 + i) 28 4 39 2 (1 + i)(2 − 2i) +1= +1= +1= + i+1= + i (3 + i)(2 − i) 7−i (7 − i)(7 + i) 50 50 25 25
(You can do this in lots of other ways — the final answer should be the same whichever order you use.) 4. Calculate (i) Comment on your answers. (i)
1 + 3i 1−i
(ii)
1 − 3i 1+i
1 + 3i (1 + 3i) (1 + i) −2 + 4i = = = −1 + 2i 1−i (1 − i) (1 + i) 2
1 − 3i (1 − 3i) (1 − i) −2 − 4i = = = −1 − 2i 1+i (1 + i) (1 − i) 2 The numbers in the initial fractions are complex conjugates of each other, so the final answers should be a pair of complex conjugate numbers. (ii)
5. Given that z = (0.6 + 0.8 i) calculate (i) z 2
(ii) z 3
(iii) z 4
and plot the original number z, and its powers z 2 , z 3 , z 4 on an Argand diagram. z = (0.6 + 0.8 i),
z 2 = (−0.28 + 0.96 i),
z 3 = (−0.936 + 0.352 i),
z 4 = (−0.8432 − 0.5376 i)
The points are on the unit circle, each separated from its neighbours by the same angle.