1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

ANSWERS 585

Section 1.5. The Algebra of Complex Numbers, Revisited: page 37

la. - 16 - il6J3.

le. -64.

3. cos39 = cos^3 9 - 3 cos 9 sin^2 9, sin 39 = 3 cos^2 8 - sin^3 8.

5a. v'2cos(~ +^2 ~") +iv'2sin (~ +^2 ~") fork= O, 1, 2.

5c. 2 ± i and -2 ± i.

5e. 2cos (~ + ~") + i2sin (~ + ~") fork= 0, 1, 2, 3.

7. Since the coefficients are real, roots come in conjugates. Thus, z - i and z + i
   are factors. Dividing the polynomial by z^2 + 1 yields a quadratic, which can
   be solved with the quadratic formula (1.5) to get 2 - i and 2 + i for the
   remaining solutions.


8. 2J3 + 2i, -4i, -2J3 + 2i.

Ha. Verify that (1 - z)(l + z + z^2 + · · · + zn) = 1 - z"+l.

llb. Let z = ei^8 For the left-hand side of part (a), use De Moivre's formula. Keep
z = e'^9 on the right-hand s ide and multiply numerator and denominator by
e-i~. Simplify, and then equate real parts of the left- and right-hand sides.

13. 
    

    Use exercise (11) and recall that if Zk is an n^1 " root of unity, then zk = 1.

    
    


14. 
    

    The four roots are ± 1 ± i (show the details). Use the roots as linear factors
    in conjugate pairs to get z^4 + 1 = (z^2 + 2z + 2) (z^2 - 2z + 2).

    
    

Section 1.6. The Topology of Complex Numbers: page 45

la. z (t) = t +it for 0 $ t $ 1.

le. z (t) = t + i for 0 $ t $ 1.

3a. z(t) = t+it^2 for 0 $ t $ 2.

3c. z ( t) = 1 - t + i (1 - t)

2

for 0 $ t $ 1.

5a. z(t) = cost+isint for - 2 " $ t $ ~·

7a. z (t) =cost+ isint for 0 $ t $ 7r/2.

9b. Open: (i), (iv), (v), (vi), and (vii). Connected: (i)-(vi). Domains: (i), (iv),

(v), and (vi). Regions: (i)-(vi). Closed regions: (iii) Bounded : (iii). (v),

and (vii).

11. Let R = Max {!z1] , lzil , ... , lznl}. Clearly, S i;;; DR (0). Thus, Sis bounded.