Solutions for Math 311 Assignment #1

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1 Solutions for Math 311 Assignment #1 (1) Show that (a) Re(iz) Im(z); (b) Im(iz) Re(z). Proof. Let z x + yi with x Re(z) and y Im(z). Then Re(iz) Re( y + xi) y Im(z) and Im(iz) Im( y + xi) x Re(z). () Verify the associative law for multiplication of complex numbers. That is, show that for all z 1, z, z 3 C. (z 1 z )z 3 z 1 (z z 3 ) Proof. Let z k x k + iy k for k 1,, 3. Then (z 1 z )z 3 ((x 1 + y 1 i)(x + y i))(x 3 + y 3 i) ((x 1 x y 1 y ) + i(x y 1 + x 1 y ))(x 3 + y 3 i) (x 1 x x 3 x 3 y 1 y x y 1 y 3 x 1 y y 3 ) + i(x x 3 y 1 + x 1 x 3 y + x 1 x y 3 y 1 y y 3 ) and z 1 (z z 3 ) (x 1 + y 1 i)((x + y i))(x 3 + y 3 i)) Therefore, (x 1 + y 1 i)((x x 3 y y 3 ) + i(x y 3 + x 3 y )) (x 1 x x 3 x 3 y 1 y x y 1 y 3 x 1 y y 3 ) + i(x x 3 y 1 + x 1 x 3 y + x 1 x y 3 y 1 y y 3 ) (z 1 z )z 3 z 1 (z z 3 ) (3) Compute (a) 1 + i 1 i ; (b) (1 i) 4. Answer. (a) i (b) 4 (4) In each case, sketch the set of points on the complex plane determined by: (a) z i ; 1

2 (b) < z i < 3; (c) z 1 z i ; (d) z + z ; (e) Im(z + i) > 3. Answer. (a) The circle x + (y 1) 4. (b) The annulus 4 (x + 3) + (y + 1) < 9. (c) The line x + 4y (d) The hyperbola x y 1. (e) The union of two half planes {y > 4} and {y < }. (5) Show that Im(z 1 z ) z 3 + z 4 z 1 + z z 3 z 4 for all complex numbers z 1, z, z 3, z 4 satisfying z 3 z 4. Also give the sufficient and necessary condition under which the equality holds. Proof. Since and we have Im(z 1 z ) z 3 + z 4 Im(z 1 z ) z 1 z z 1 + z z 3 + z 4 z3 z 4 Im(z 1 z ) z 3 + z 4 z 1 + z z 3 z 4 z 1 + z z 3 z 4. If the equality holds, we must have Re(z 1 z ) 0, Im(z 1 z ) 0 and z 1 z z 1 + z. It follows that x 1 x 0, y 1 y and y 1 y 0, where we let z k x k + iy k for k 1,, 3, 4. If z 1 0 or z 0, we also have z 3 + z 4 z3 z 4, i.e., z 3 λz 4 or z 4 λz 3 for some λ < 0. In conclusion, the inequality holds if x 1 x 0, y 1 y, y 1 y 0 and z 1 z 0 or z 3 λz 4 or z 4 λz 3 for some λ < 0. (6) Use the properties of conjugates and moduli to show that (a) z + 3i z 3i; (b) iz iz; (c) (z + 5)( i) 3 z + 5.

3 3 Proof. (a) z + 3i z + 3i z 3i. (b) iz iz iz (c) (z + 5)( i) z + 5 i 3 z z + 5 Solutions for Math 311 Assignment # (1) Find the principal argument Arg(z) when (a) z i 1 + i ; (b) z ( 3 + i) 010. Solution. (a) ( ) i Arg Arg 1 + i (b) Arg(( 3 + i) 010 ) Arg ( 1 + i ) ( ) Arg eπi/4 π/4 ( ) i 010 Arg((e πi/6 ) 010 ) Arg(e 335πi ) Arg(e πi ) π () Show that if Re(z 1 ) > 0 and Re(z ) > 0, then Proof. We always have Arg(z 1 z ) Arg(z 1 ) + Arg(z ). Arg(z 1 z ) Arg(z 1 ) + Arg(z ) + kπ for some integer k. Since Re(z 1 ) > 0 and Re(z ) > 0, π/ < Arg(z 1 ) < π/ and π/ < Arg(z ) < π/ and hence And since π < Arg(z 1 ) + Arg(z ) < π. π < Arg(z 1 z ) π,

4 4 π < Arg(z 1 z ) (Arg(z 1 ) + Arg(z )) < π Therefore, π < kπ < π and 1 < k < 1. Hence k 0. (3) Use the identity to derive sin(kθ) k1 k0 z k 1 zn+1 1 z cos(θ/) cos((n + 1)θ/) sin(θ/) Proof. Let z e iθ. Then cos(kθ) + i sin(kθ) k0 k1 k0 e ikθ 1 ei(n+1)θ 1 e iθ 1 e i(n+1)θ (1 cos θ) i sin θ 1 e i(n+1)θ sin(θ/)(sin(θ/) i cos(θ/)) 1 e i(n+1)θ sin(θ/)( i)e i(e θi/ e (n+1)iθ/ ) θi/ sin(θ/) (sin((n + 1)θ/) + sin(θ/)) + i(cos(θ/) cos((n + 1)θ/) sin(θ/) Therefore, sin(kθ) k1 cos(θ/) cos((n + 1)θ/. sin(θ/) (4) Find all the square roots of (a) 1 + i and (b) 1 + 3i. Express them in rectangular coordinates. Solution. (a) 1 + i e πi/4 4 e kπi+πi/8 ± 4 e πi/8 By half-angle formula, cos(π/8) 4 + / and sin(π/8) 4 /. Therefore, 1 + i ± 4 ( i 4 )

5 (b) 1 + 3i e πi/3 e kπi+πi/6 ± ( ) 6 e πi/6 ± + i 5 (5) Find the four zeros of the polynomial z and use these to factor z into quadratic factors with real coefficients. Solution. The 4th roots of 4 are e πi ( kπi exp + πi ) 4 for k 0, 1,, 3. So z (z e πi/4 )(z e 3πi/4 )(z e 5πi/4 )(z e 7πi/4 ) (z e πi/4 )(z e πi/4 )(z e 3πi/4 )(z e 3πi/4 ) (z (e πi/4 + e πi/4 )z + )(z (e 3πi/4 + e 3πi/4 )z + ) (z z + )(z + z + ) (6) Sketch the following sets and determine which are domains, which are closed and which are bounded: (a) z + i 1; (b) z + 3 > 4; (c) Im(z) > 1; (d) z 4 z ; (e) Im(z) 1; (f) 0 arg(z) π/4(z 0). Justify your answer. Solution. (a) It is not a domain since it is not open; choose a point z 0 satisfying z 0 +i 1 and the disk z z 0 < r is not contained in the set for all r > 0. It is closed since { z +i > 1} is open and it is bounded since z i + 1 < 4 for all z satisfying z + i 1. (b) It is a domain since it is open and connected. It is not closed since { z + 3 4} is not open and it is not bounded since one can find a sequence z n satisfying z n + 3 > 4 such that lim n z n. For example, let z n n. (c) It is a domain since it is open and connected. It is not closed since {Im(z) 1} is not open and it is not bounded since one can find a sequence z n satisfying Im(z) > 1 such that lim n z n. For example, let z n ni.

6 6 (d) It is not a domain since it is not open; choose a point z 0 satisfying z 4 z and the disk z z 0 < r is not contained in the set for all r > 0. It is closed since { z 4 < z } is open. It is not bounded since one can find a sequence z n satisfying z 4 z such that lim n z n. For example, let z n ni. (e) It is not a domain since it is not open; choose a point z 0 satisfying Im(z) 1 and the disk z z 0 < r is not contained in the set for all r > 0. It is closed since {Im(z) 1} is open. It is not bounded since one can find a sequence z n satisfying Im(z) 1 such that lim n z n. For example, let z n n + i. (f) It is not a domain since it is not open; choose a point z 0 satisfying arg(z) 0 and the disk z z 0 < r is not contained in the set for all r > 0. It is not closed since 0 does not lie in the set and the disk z < r meets the set for all r > 0. It is not bounded since one can find a sequence z n satisfying arg(z) 0 such that lim n z n. For example, let z n n. (7) Write the following functions f(z) in the forms f(z) u(x, y)+ iv(x, y) under Cartesian coordinates and f(z) u(r, θ)+iv(r, θ) under polar coordinates: (a) f(z) z + z + 1; (b) f(z) 1 + z 1 z. Solution. (a) f(z) (x + yi) + (x + yi) + 1 (x y + x + 1) + i(xy + y) r e iθ + re iθ + 1 (r cos(θ) + r cos θ + 1) + i(r sin(θ) + r sin θ) (b) f(z) (1 + z)(1 z) (1 z)(1 z) 1 + iy (x + y ) (1 x) + y 1 (x + y ) (1 x) + y + i y (1 x) + y 1 r 1 + r r cos θ + i r sin θ 1 + r r cos θ

7 (8) Sketch the region onto which the sector {r 1, 0 θ π/4} is mapped by the transformation (a) w z 3 and (b) w z 4. Solution. (a) The image is { w 1, 0 arg(w) 3π/4}. (b) The image is { w 1, 0 arg(w) π}. 7