lina-09-10-09-s1 - Linear Algebra & Geometry:...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Linear Algebra & Geometry: solutions to sheet 2 1. ( a) 1 1 = √ 2u(π/4) (b) 0 −5 = 5u(3π/2) (c) −3 4 = 5u(θ) , where θ = π/2 + arccos(4/5) = 2.214 . . .. 2. (a) 1 2 1 √ 3 (b) 0 2 √ − 3/2 (c) 10 −1/2 √ = −5 3 1 3. Let v1 , v2 , · · · , vn ∈ R2 be n arbitrary vectors in R2 . We apply the triangle inequality n − 1 times v1 + v2 + · · · + vn ≤ v1 + v2 + v3 + · · · + vn ≤ v1 + v2 + v3 + v4 + · · · + vn ··· ≤ v1 + v2 + · · · + vn Equality holds for instance if v1 = v2 = v3 = · · · = vn . 4. (a) (1 + i) + (2 − 3i) = 3 − 2i (b) (1 + i)(1 + 2i) − 3 = 1 − 2 + 3i − 3 = −4 + 3i (c) (c) (e) (f) 5. 1 1−i 2+i 1+i = = 1+i (1−i)(1+i) (2+i)(1−i) (1+i)(1−i) = = 1+i 2 3−i 2 = = 1 2 3 2 + 1i 2 1 − 2i (2−3i)(3−2i) 1+i 1 (1+i)(2−i) = −13i 1+i = −13i(1−i) (1+i)(1−i) = −13−13i 2 = − 13 − 2 13 i 2 = 1 3+i = 3−i 9+1 = 3 10 − 1 i 10 (i) (a) We have z1 + z2 = x1 + x2 + i(y1 + y2 ) and so z1 + z2 = x1 + x2 − i(y1 + y2 ) = x1 − iy1 + x2 − iy2 = z1 + z2 ¯ ¯ (b) We have z1 z2 = (x1 − iy1 )(x2 − iy2 ) = x1 x2 − y1 y2 − i(x1 y2 + x2 y1 ) and ¯¯ z1 z2 = x1 x2 − y1 y2 + i(x1 y2 + x2 y1 ) = x1 x2 − y1 y2 − i(x1 y2 + x2 y1 ) (c) we have z1 /z2 = (z1 z2 )/(z2 z2 ) = (z1 z2 )/|z2 |2 and using (b) and (d) we get ¯ ¯ ¯ z1 /z2 = z1 z2 /|z2 |2 = z1 /z2 ¯ ¯¯ ¯ (d) z = x − iy = x + iy = z √ √ (ii) (e) |z1 z2 | = z1 z2 z1 z2 = z1 z1 z2 z2 = |z1 ||z2 | ¯¯ ¯¯ x1 x2 (f) With v1 = and v2 = we have v1 = |z1 |, v2 = |z2 | and y1 y2 v1 + v2 = |z1 + z2 | and so the triangle inequality in R2 gives |z1 + z2 | ≤ |z1 | + |z2 | 1 6. (i) (a) ei 3 = cos π/3 + i sin π/3 = 1 + 23 i 2 π (b) 2ei 2 = 2 cos π/2 + 2i sin π/2 = 2i (ii) (a) 1 + i = √ iπ/4 2e (b) π √ √ 7π (c) 10ei 6 = 10 cos 6π/6 + 10i sin 7π/6 = −5 3 − 5i − 5i = 5eiπ (b) − 3 + 4i = 5eiϕ where ϕ = π/2 + arccos(4/5) = 2.214 . . .. 2 ...
View Full Document

Page1 / 2

lina-09-10-09-s1 - Linear Algebra & Geometry:...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online