584 ANSWERS25c. Letting z1 = 2i, and z2 = - 2i, we compute K = 13 + 2i - 2i l + 13 + 2i + 2il =
3 + 5 = 8. Then, with z = (x, y), the equation in Exercise 25 a becomes
J x2 + (y - 2)^2 + J x^2 + (y + 2)2 = 8. Show the details that squaring both
sides, simplifying, squaring again, and simplifying again gives 4x^2 + 3y^2 = 48.
In standard form, x^2 + ~y^2 = 12.
Section 1.4. The Geometry of Complex Numbers, Continued: page
29la. -l
l e. 2 !-
le. -l
lg. - l
3a. 4 (cos11' + i sin 11') = 4ei.-.3c. 7 (cos-,/'+ isin -.;") = 7 e-~.
3e. ~(cos~+ isin ~) = ~ei~.3g. 5 (cos 8 + i sin IJ) = 5ei^9 , where IJ = Arc tan~.
5a. i.5c. 4 + i4v'3.
5e. v'2-ivlz.
5g. - e^2.7. When z = v'3 + i, Arg (iz) = Arg(z) + ~' Arg(- z) = Arg(z)- rr,
Arg (-iz) = Arg (z) - I·
- The negative real numbers and the number zero. Prove this!
11. Let IJ E arg( ~ ). Then ~ = rei^9. Hence, z = ~e-i^8 , so -IJ E arg z , or
IJ E - arg z. Thus, arg ( ~) <;;; - arg z. T he proof that -arg z <;;; arg( ~) is
similar.13a. Let 0 ~ z = x + iy. Since zz = x^2 + y^2 > 0, Arg ( zz) = 0.
15 '. From the figure it is clear that Arg(z - zo) = ¢, and lz - zol = p. The
exponential form for z - z 0 then gives the desired conclusion.