11.7E: Exercises

In Exercises 1 - 20, find a polar representation for the complex number \(z\) and then identify \(\operatorname{Re}(z)\), \(\operatorname{Im}(z),|z|, \arg (z)\) and \(\operatorname{Arg}(z)\).

1. \(z=9+9 i\)
2. \(z=5+5 i \sqrt{3}\)
3. \(z = 6i\)
4. \(z=-3 \sqrt{2}+3 i \sqrt{2}\)
5. \(z=-6 \sqrt{3}+6 i\)
6. \(z = −2\)
7. \(z=-\frac{\sqrt{3}}{2}-\frac{1}{2} i\)
8. \(z = −3 − 3i\)
9. \(z = −5i\)
10. \(z=2 \sqrt{2}-2 i \sqrt{2}\)
11. \(z = 6\)
12. \(z=i \sqrt[3]{7}\)
13. \(z = 3 + 4i\)
14. \(z=\sqrt{2}+i\)
15. \(z = −7 + 24i\)
16. \(z = −2 + 6i\)
17. \(z = −12 − 5i\)
18. \(z = −5 − 2i\)
19. \(z = 4 − 2i\)
20. \(z = 1 − 3i\)

In Exercises 21 - 40, find the rectangular form of the given complex number. Use whatever identities are necessary to find the exact values.

21. \(z=6 \operatorname{cis}(0)\)
22. \(z=2 \operatorname{cis}\left(\frac{\pi}{6}\right)\)
23. \(z=7 \sqrt{2} \operatorname{cis}\left(\frac{\pi}{4}\right)\)
24. \(z=3 \operatorname{cis}\left(\frac{\pi}{2}\right)\)
25. \(z=4 \operatorname{cis}\left(\frac{2 \pi}{3}\right)\)
26. \(z=\sqrt{6} \operatorname{cis}\left(\frac{3 \pi}{4}\right)\)
27. \(z=9 \operatorname{cis}(\pi)\)
28. \(z=3 \operatorname{cis}\left(\frac{4 \pi}{3}\right)\)
29. \(z=7 \operatorname{cis}\left(-\frac{3 \pi}{4}\right)\)
30. \(z=\sqrt{13} \operatorname{cis}\left(\frac{3 \pi}{2}\right)\)
31. \(z=\frac{1}{2} \operatorname{cis}\left(\frac{7 \pi}{4}\right)\)
32. \(z=12 \operatorname{cis}\left(-\frac{\pi}{3}\right)\)
33. \(z=8 \operatorname{cis}\left(\frac{\pi}{12}\right)\)
34. \(z=2 \operatorname{cis}\left(\frac{7 \pi}{8}\right)\)
35. \(z=5 \operatorname{cis}\left(\arctan \left(\frac{4}{3}\right)\right)\)
36. \(z=\sqrt{10} \operatorname{cis}\left(\arctan \left(\frac{1}{3}\right)\right)\)
37. \(z=15 \operatorname{cis}(\arctan (-2))\)
38. \(z=\sqrt{3}(\arctan (-\sqrt{2}))\)
39. \(z=50 \operatorname{cis}\left(\pi-\arctan \left(\frac{7}{24}\right)\right)\)
40. \(z=\frac{1}{2} \operatorname{cis}\left(\pi+\arctan \left(\frac{5}{12}\right)\right)\)

For Exercises 41 - 52, use \(z=-\frac{3 \sqrt{3}}{2}+\frac{3}{2} i\) and \(w=3 \sqrt{2}-3 i \sqrt{2}\) to compute the quantity. Express your answers in polar form using the principal argument.

41. \(zw\)
42. \(\frac{z}{w}\)
43. \(\frac{w}{z}\)
44. \(z^{4}\)
45. \(w^{3}\)
46. \(z^{5} w^{2}\)
47. \(z^{3} w^{2}\)
48. \(\frac{z^{2}}{w}\)
49. \(\frac{w}{z^{2}}\)
50. \(\frac{z^{3}}{w^{2}}\)
51. \(\frac{w^{2}}{z^{3}}\)
52. \(\left(\frac{w}{z}\right)^{6}\)

In Exercises 53 - 64, use DeMoivre’s Theorem to find the indicated power of the given complex number. Express your final answers in rectangular form.

53. \((-2+2 i \sqrt{3})^{3}\)
54. \((-\sqrt{3}-i)^{3}\)
55. \((-3+3 i)^{4}\)
56. \((\sqrt{3}+i)^{4}\)
57. \(\left(\frac{5}{2}+\frac{5}{2} i\right)^{3}\)
58. \(\left(-\frac{1}{2}-\frac{\sqrt{3}}{2} i\right)^{6}\)
59. \(\left(\frac{3}{2}-\frac{3}{2} i\right)^{3}\)
60. \(\left(\frac{\sqrt{3}}{3}-\frac{1}{3} i\right)^{4}\)
61. \(\left(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2} i\right)^{4}\)
62. \((2+2 i)^{5}\)
63. \((\sqrt{3}-i)^{5}\)
64. \((1-i)^{8}\)

In Exercises 65 - 76, find the indicated complex roots. Express your answers in polar form and then convert them into rectangular form.

65. the two square roots of \(z = 4i\)
66. the two square roots of \(z = −25i\)
67. the two square roots of \(z=1+i \sqrt{3}\)
68. the two square roots of \(\frac{5}{2}-\frac{5 \sqrt{3}}{2} i\)
69. the three cube roots of \(z = 64\)
70. the three cube roots of \(z = −125\)
71. the three cube roots of \(z = i\)
72. the three cube roots of \(z = −8i\)
73. the four fourth roots of \(z = 16\)
74. the four fourth roots of \(z = −81\)
75. the six sixth roots of \(z = 64\)
76. the six sixth roots of \(z = −729\)
77. Use the Sum and Difference Identities in Theorem 10.16 or the Half Angle Identities in Theorem 10.19 to express the three cube roots of \(z=\sqrt{2}+i \sqrt{2}\) in rectangular form. (See Example 11.7.4, number 3.)
78. Use a calculator to approximate the five fifth roots of 1. (See Example 11.7.4, number 4.)
79. According to Theorem 3.16 in Section 3.4, the polynomial \(p(x)=x^{4}+4\) can be factored into the product linear and irreducible quadratic factors. In Exercise 28 in Section 8.7, we showed you how to factor this polynomial into the product of two irreducible quadratic factors using a system of non-linear equations. Now that we can compute the complex fourth roots of −4 directly, we can simply apply the Complex Factorization Theorem, Theorem 3.14, to obtain the linear factorization \(p(x) = (x − (1 + i))(x − (1 − i))(x − (−1 + i))(x − (−1 − i))\). By multiplying the first two factors together and then the second two factors together, thus pairing up the complex conjugate pairs of zeros Theorem 3.15 told us we’d get, we have that \(p(x)=\left(x^{2}-2 x+2\right)\left(x^{2}+2 x+2\right)\). Use the 12 complex 12^th roots of 4096 to factor \(p(x)=x^{12}-4096\) into a product of linear and irreducible quadratic factors.
80. Complete the proof of Theorem 11.14 by showing that if \(w \neq 0\) than \(\left|\frac{1}{w}\right|=\frac{1}{|w|}\).
81. Recall from Section 3.4 that given a complex number \(z = a+bi\) its complex conjugate, denoted \(\bar{z}\), is given by \(\bar{z}=a-b i\).
    1. Prove that \(|\bar{z}|=|z|\).
    2. Prove that \(|z|=\sqrt{z \bar{z}}\)
    3. Show that \(\operatorname{Re}(z)=\frac{z+\bar{z}}{2}\) and \(\operatorname{Im}(z)=\frac{z-\bar{z}}{2 i}\)
    4. Show that if \(\theta \in \arg (z)\) then \(-\theta \in \arg (\bar{z})\). Interpret this result geometrically.
    5. Is it always true that \(\operatorname{Arg}(\bar{z})=-\operatorname{Arg}(z)\)?
82. Given any natural number \(n \geq 2\), the \(n\) complex \(n^{\text {th }}\) roots of the number \(z = 1\) are called the \(n^{\text {th }}\) Roots of Unity. In the following exercises, assume that \(n\) is a fixed, but arbitrary, natural number such that \(n \geq 2\).
    1. Show that \(w = 1\) is an \(n^{\text {th }}\) root of unity.
    2. Show that if both \(w_{j}\) and \(w_{k}\) are \(n^{\text {th }}\) roots of unity then so is their product \(w_{j} w_{k}\).
    3. Show that if \(w_{j}\) is an \(n^{\text {th }}\) root of unity then there exists another \(n^{\text {th }}\) root of unity \(w_{j^{\prime}}\) such that \(w_{j} w_{j^{\prime}}=1\). Hint: If \(w_{j}=\operatorname{cis}(\theta)\) let \(w_{j^{\prime}}=\operatorname{cis}(2 \pi-\theta)\). You’ll need to verify that \(w_{j^{\prime}}=\operatorname{cis}(2 \pi-\theta)\) is indeed an \(n^{\text {th }}\) root of unity.
83. Another way to express the polar form of a complex number is to use the exponential function. For real numbers \(t\), Euler’s Formula defines \(e^{i t}=\cos (t)+i \sin (t)\).
    1. Use Theorem 11.16 to show that \(e^{i x} e^{i y}=e^{i(x+y)}\) for all real numbers \(x\) and \(y\).
    2. Use Theorem 11.16 to show that \(\left(e^{i x}\right)^{n}=e^{i(n x)}\) for any real number \(x\) and any natural number \(n\).
    3. Use Theorem 11.16 to show that \(\frac{e^{i x}}{e^{i y}}=e^{i(x-y)}\) for all real numbers \(x\) and \(y\).
    4. If \(z=r \operatorname{cis}(\theta)\) is the polar form of \(z\), show that \(z=r e^{i t}\) where \(\theta=t\) radians.
    5. Show that \(e^{i \pi}+1=0\). (This famous equation relates the five most important constants in all of Mathematics with the three most fundamental operations in Mathematics.)
    6. Show that \(\cos (t)=\frac{e^{i t}+e^{-i t}}{2}\) and that \(\sin (t)=\frac{e^{i t}-e^{-i t}}{2 i}\) for all real numbers \(t\).

11.7.2 Answers

1. \(\begin{aligned}
   &z=9+9 i=9 \sqrt{2} \operatorname{cis}\left(\frac{\pi}{4}\right), \quad \operatorname{Re}(z)=9, \quad \operatorname{Im}(z)=9, \quad|z|=9 \sqrt{2} \\
   &\arg (z)=\left\{\frac{\pi}{4}+2 \pi k \mid k \text { is an integer }\right\} \text { and } \operatorname{Arg}(z)=\frac{\pi}{4}.
   \end{aligned}\)
2. \(\begin{aligned}
   &z=5+5 i \sqrt{3}=10 \operatorname{cis}\left(\frac{\pi}{3}\right), \quad \operatorname{Re}(z)=5, \quad \operatorname{Im}(z)=5 \sqrt{3}, \quad|z|=10 \\
   &\arg (z)=\left\{\frac{\pi}{3}+2 \pi k \mid k \text { is an integer }\right\} \text { and } \operatorname{Arg}(z)=\frac{\pi}{3}
   \end{aligned}\)
3. \(\begin{aligned}
   &z=6 i=6 \operatorname{cis}\left(\frac{\pi}{2}\right), \quad \operatorname{Re}(z)=0, \quad \operatorname{Im}(z)=6, \quad|z|=6 \\
   &\arg (z)=\left\{\frac{\pi}{2}+2 \pi k \mid k \text { is an integer }\right\} \text { and } \operatorname{Arg}(z)=\frac{\pi}{2}.
   \end{aligned}\)
4. \(\begin{aligned}
   &z=-3 \sqrt{2}+3 i \sqrt{2}=6 \operatorname{cis}\left(\frac{3 \pi}{4}\right), \quad \operatorname{Re}(z)=-3 \sqrt{2}, \quad \operatorname{Im}(z)=3 \sqrt{2}, \quad|z|=6 \\
   &\arg (z)=\left\{\frac{3\pi}{2}+2 \pi k \mid k \text { is an integer }\right\} \text { and } \operatorname{Arg}(z)=\frac{3\pi}{2}.
   \end{aligned}\)
5. \(\begin{aligned}
   &z=-6 \sqrt{3}+6 i=12 \operatorname{cis}\left(\frac{5 \pi}{6}\right), \quad \operatorname{Re}(z)=-6 \sqrt{3}, \quad \operatorname{Im}(z)=6, \quad|z|=12 \\
   &\arg (z)=\left\{\frac{5\pi}{2}+2 \pi k \mid k \text { is an integer }\right\} \text { and } \operatorname{Arg}(z)=\frac{5\pi}{2}.
   \end{aligned}\)
6. \(\begin{aligned}
   &z=-2=2 \operatorname{cis}(\pi), \quad \operatorname{Re}(z)=-2, \quad \operatorname{Im}(z)=0, \quad|z|=2\\
   &\arg (z)=\left\{(2k+1)\pi \mid k \text { is an integer }\right\} \text { and } \operatorname{Arg}(z)=\pi.
   \end{aligned}\)
7. \(\begin{aligned}
   &z=-\frac{\sqrt{3}}{2}-\frac{1}{2} i=\operatorname{cis}\left(\frac{7 \pi}{6}\right), \quad \operatorname{Re}(z)=-\frac{\sqrt{3}}{2}, \quad \operatorname{Im}(z)=-\frac{1}{2}, \quad|z|=1 \\
   &\arg (z)=\left\{\frac{7\pi}{2}+2 \pi k \mid k \text { is an integer }\right\} \text { and } \operatorname{Arg}(z)=-\frac{5\pi}{2}.
   \end{aligned}\)
8. \(\begin{aligned}
   &z=-3-3 i=3 \sqrt{2} \operatorname{cis}\left(\frac{5 \pi}{4}\right), \quad \operatorname{Re}(z)=-3, \quad \operatorname{Im}(z)=-3, \quad|z|=3 \sqrt{2} \\
   &\arg (z)=\left\{\frac{5\pi}{4}+2 \pi k \mid k \text { is an integer }\right\} \text { and } \operatorname{Arg}(z)=-\frac{3\pi}{4}.
   \end{aligned}\)
9. \(\begin{aligned}
   &z=-5 i=5 \operatorname{cis}\left(\frac{3 \pi}{2}\right), \quad \operatorname{Re}(z)=0, \quad \operatorname{Im}(z)=-5, \quad|z|=5 \\
   &\arg (z)=\left\{\frac{3\pi}{2}+2 \pi k \mid k \text { is an integer }\right\} \text { and } \operatorname{Arg}(z)=-\frac{\pi}{2}.
   \end{aligned}\)
10. \(\begin{aligned}
    &z=2 \sqrt{2}-2 i \sqrt{2}=4 \operatorname{cis}\left(\frac{7 \pi}{4}\right), \quad \operatorname{Re}(z)=2 \sqrt{2}, \quad \operatorname{Im}(z)=-2 \sqrt{2}, \quad|z|=4 \\
    &\arg (z)=\left\{\frac{7\pi}{4}+2 \pi k \mid k \text { is an integer }\right\} \text { and } \operatorname{Arg}(z)=-\frac{\pi}{4}.
    \end{aligned}\)
11. \(\begin{aligned}
    &z=6=6 \operatorname{cis}(0), \quad \operatorname{Re}(z)=6, \quad \operatorname{Im}(z)=0, \quad|z|=6 \\
    &\arg (z)=\left\{2 \pi k \mid k \text { is an integer }\right\} \text { and } \operatorname{Arg}(z)=0.
    \end{aligned}\)
12. \(\begin{aligned}
    &z=i \sqrt[3]{7}=\sqrt[3]{7} \operatorname{cis}\left(\frac{\pi}{2}\right), \quad \operatorname{Re}(z)=0, \quad \operatorname{Im}(z)=\sqrt[3]{7}, \quad|z|=\sqrt[3]{7}
    \\ &\arg (z)=\left\{\frac{\pi}{2}+2 \pi k \mid k \text { is an integer }\right\} \text { and } \operatorname{Arg}(z)=\frac{\pi}{2}.
    \end{aligned}\)
13. \(\begin{aligned}
    &z=3+4 i=5 \operatorname{cis}\left(\arctan \left(\frac{4}{3}\right)\right), \quad \operatorname{Re}(z)=3, \quad \operatorname{Im}(z)=4, \quad|z|=5
    \\ &\arg (z)=\left\{\operatorname{arctan}\left(\frac{4}{3}\right)+2 \pi k \mid k \text { is an integer }\right\} \text { and } \operatorname{Arg}(z)=\operatorname{arctan}\left(\frac{4}{3}\right).
    \end{aligned}\)
14. \(\begin{aligned}
    &z=\sqrt{2}+i=\sqrt{3} \operatorname{cis}\left(\arctan \left(\frac{\sqrt{2}}{2}\right)\right), \quad \operatorname{Re}(z)=\sqrt{2}, \quad \operatorname{Im}(z)=1, \quad|z|=\sqrt{3}
    \\ &\arg (z)=\left\{\operatorname{arctan}\left(\frac{\sqrt{2}}{2}\right)+2 \pi k \mid k \text { is an integer }\right\} \text { and } \operatorname{Arg}(z)=\operatorname{arctan}\left(\frac{\sqrt{2}}{2}\right).
    \end{aligned}\)
15. \(\begin{aligned}
    &z=-7+24 i=25 \operatorname{cis}\left(\pi-\arctan \left(\frac{24}{7}\right)\right), \quad \operatorname{Re}(z)=-7, \quad \operatorname{Im}(z)=24, \quad|z|=25
    \\ &\arg (z)=\left\{\pi-\operatorname{arctan}\left(\frac{24}{7}\right)+2 \pi k \mid k \text { is an integer }\right\} \text { and } \operatorname{Arg}(z)=\pi-\operatorname{arctan}\left(\frac{24}{7}\right).
    \end{aligned}\)
16. \(\begin{aligned}
    &z=-2+6 i=2 \sqrt{10} \operatorname{cis}(\pi-\arctan (3)), \quad \operatorname{Re}(z)=-2, \quad \operatorname{Im}(z)=6, \quad|z|=2 \sqrt{10}
    \\ &\arg (z)=\left\{\pi-\operatorname{arctan}(3)+2 \pi k \mid k \text { is an integer }\right\} \text { and } \operatorname{Arg}(z)=\pi-\operatorname{arctan}(3).
    \end{aligned}\)
17. \(\begin{aligned}
    &z=-12-5 i=13 \operatorname{cis}\left(\pi+\arctan \left(\frac{5}{12}\right)\right), \quad \operatorname{Re}(z)=-12, \quad \operatorname{Im}(z)=-5, \quad|z|=13
    \\ &\arg (z)=\left\{\pi+\operatorname{arctan}\left(\frac{5}{12}\right)+2 \pi k \mid k \text { is an integer }\right\} \text { and } \operatorname{Arg}(z)=\operatorname{arctan}\left(\frac{5}{12}\right)-\pi.
    \end{aligned}\)
18. \(\begin{aligned}
    &z=-5-2 i=\sqrt{29} \operatorname{cis}\left(\pi+\arctan \left(\frac{2}{5}\right)\right), \quad \operatorname{Re}(z)=-5, \quad \operatorname{Im}(z)=-2, \quad|z|=\sqrt{29}
    \\ &\arg (z)=\left\{\pi+\operatorname{arctan}\left(\frac{2}{5}\right)+2 \pi k \mid k \text { is an integer }\right\} \text { and } \operatorname{Arg}(z)=\operatorname{arctan}\left(\frac{2}{5}\right)-\pi.
    \end{aligned}\)
19. \(\begin{aligned}
    &z=4-2 i=2 \sqrt{5} \operatorname{cis}\left(\arctan \left(-\frac{1}{2}\right)\right), \quad \operatorname{Re}(z)=4, \quad \operatorname{Im}(z)=-2, \quad|z|=2 \sqrt{5}
    \\ &\arg (z)=\left\{\operatorname{arctan}\left(-\frac{1}{2}\right)+2 \pi k \mid k \text { is an integer }\right\} \text { and } \operatorname{Arg}(z)=\operatorname{arctan}\left(-\frac{1}{2}\right)=-\operatorname{arctan}\left(\frac{1}{2}\right).
    \end{aligned}\)
20. \(\begin{aligned}
    &z=1-3 i=\sqrt{10} \operatorname{cis}(\arctan (-3)), \quad \operatorname{Re}(z)=1, \quad \operatorname{Im}(z)=-3, \quad|z|=\sqrt{10}
    \\ &\arg (z)=\left\{\operatorname{arctan}(-3)+2 \pi k \mid k \text { is an integer }\right\} \text { and } \operatorname{Arg}(z)=\operatorname{arctan}(-3)=-\operatorname{arctan}(3).
    \end{aligned}\)
21. \(z=6 \operatorname{cis}(0)=6\)
22. \(z=2 \operatorname{cis}\left(\frac{\pi}{6}\right)=\sqrt{3}+i\)
23. \(z=7 \sqrt{2} \operatorname{cis}\left(\frac{\pi}{4}\right)=7+7 i\)
24. \(z=3 \operatorname{cis}\left(\frac{\pi}{2}\right)=3 i\)
25. \(z=4 \operatorname{cis}\left(\frac{2 \pi}{3}\right)=-2+2 i \sqrt{3}\)
26. \(z=\sqrt{6} \operatorname{cis}\left(\frac{3 \pi}{4}\right)=-\sqrt{3}+i \sqrt{3}\)
27. \(z=9 \operatorname{cis}(\pi)=-9\)
28. \(z=3 \operatorname{cis}\left(\frac{4 \pi}{3}\right)=-\frac{3}{2}-\frac{3 i \sqrt{3}}{2}\)
29. \(z=7 \operatorname{cis}\left(-\frac{3 \pi}{4}\right)=-\frac{7 \sqrt{2}}{2}-\frac{7 \sqrt{2}}{2} i\)
30. \(z=\sqrt{13} \operatorname{cis}\left(\frac{3 \pi}{2}\right)=-i \sqrt{13}\)
31. \(z=\frac{1}{2} \operatorname{cis}\left(\frac{7 \pi}{4}\right)=\frac{\sqrt{2}}{4}-i \frac{\sqrt{2}}{4}\)
32. \(z=12 \operatorname{cis}\left(-\frac{\pi}{3}\right)=6-6 i \sqrt{3}\)
33. \(z=8 \operatorname{cis}\left(\frac{\pi}{12}\right)=4 \sqrt{2+\sqrt{3}}+4 i \sqrt{2-\sqrt{3}}\)
34. \(z=2 \operatorname{cis}\left(\frac{7 \pi}{8}\right)=-\sqrt{2+\sqrt{2}}+i \sqrt{2-\sqrt{2}}\)
35. \(z=5 \operatorname{cis}\left(\arctan \left(\frac{4}{3}\right)\right)=3+4 i\)
36. \(z=\sqrt{10} \operatorname{cis}\left(\arctan \left(\frac{1}{3}\right)\right)=3+i\)
37. \(z=15 \operatorname{cis}(\arctan (-2))=3 \sqrt{5}-6 i \sqrt{5}\)
38. \(z=\sqrt{3} \operatorname{cis}(\arctan (-\sqrt{2}))=1-i \sqrt{2}\)
39. \(z=50 \operatorname{cis}\left(\pi-\arctan \left(\frac{7}{24}\right)\right)=-48+14 i\)
40. \(z=\frac{1}{2} \operatorname{cis}\left(\pi+\arctan \left(\frac{5}{12}\right)\right)=-\frac{6}{13}-\frac{5 i}{26}\)

In Exercises 41 - 52, we have that \(z=-\frac{3 \sqrt{3}}{2}+\frac{3}{2} i=3 \operatorname{cis}\left(\frac{5 \pi}{6}\right)\) and \(w=3 \sqrt{2}-3 i \sqrt{2}=6 \operatorname{cis}\left(-\frac{\pi}{4}\right)\) so we get the following.

41. \(z w=18 \operatorname{cis}\left(\frac{7 \pi}{12}\right)\)
42. \(\frac{z}{w}=\frac{1}{2} \operatorname{cis}\left(-\frac{11 \pi}{12}\right)\)
43. \(\frac{w}{z}=2 \operatorname{cis}\left(\frac{11 \pi}{12}\right)\)
44. \(z^{4}=81 \operatorname{cis}\left(-\frac{2 \pi}{3}\right)\)
45. \(w^{3}=216 \operatorname{cis}\left(-\frac{3 \pi}{4}\right)\)
46. \(z^{5} w^{2}=8748 \operatorname{cis}\left(-\frac{\pi}{3}\right)\)
47. \(z^{3} w^{2}=972 \operatorname{cis}(0)\)
48. \(\frac{z^{2}}{w}=\frac{3}{2} \operatorname{cis}\left(-\frac{\pi}{12}\right)\)
49. \(\frac{w}{z^{2}}=\frac{2}{3} \operatorname{cis}\left(\frac{\pi}{12}\right)\)
50. \(\frac{z^{3}}{w^{2}}=\frac{3}{4} \operatorname{cis}(\pi)\)
51. \(\frac{w^{2}}{z^{3}}=\frac{4}{3} \operatorname{cis}(\pi)\)
52. \(\left(\frac{w}{z}\right)^{6}=64 \operatorname{cis}\left(-\frac{\pi}{2}\right)\)
53. \((-2+2 i \sqrt{3})^{3}=64\)
54. \((-\sqrt{3}-i)^{3}=-8 i\)
55. \((-3+3 i)^{4}=-324\)
56. \((\sqrt{3}+i)^{4}=-8+8 i \sqrt{3}\)
57. \(\left(\frac{5}{2}+\frac{5}{2} i\right)^{3}=-\frac{125}{4}+\frac{125}{4} i\)
58. \(\left(-\frac{1}{2}-\frac{i \sqrt{3}}{2}\right)^{6}=1\)
59. \(\left(\frac{3}{2}-\frac{3}{2} i\right)^{3}=-\frac{27}{4}-\frac{27}{4} i\)
60. \(\left(\frac{\sqrt{3}}{3}-\frac{1}{3} i\right)^{4}=-\frac{8}{81}-\frac{8 i \sqrt{3}}{81}\)
61. \(\left(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2} i\right)^{4}=-1\)
62. \((2+2 i)^{5}=-128-128 i\)
63. \((\sqrt{3}-i)^{5}=-16 \sqrt{3}-16 i\)
64. \((1-i)^{8}=16\)
65. \(\begin{aligned}
    &\text{Since }z=4 i=4 \operatorname{cis}\left(\frac{\pi}{2}\right)\text{ we have} \\
    &w_{0}=2 \operatorname{cis}\left(\frac{\pi}{4}\right)=\sqrt{2}+i \sqrt{2} \quad\quad\quad\quad\quad w_{1}=2 \operatorname{cis}\left(\frac{5 \pi}{4}\right)=-\sqrt{2}-i \sqrt{2}
    \end{aligned}\)
66. \(\begin{aligned}
    &\text{Since }z=-25 i=25 \operatorname{cis}\left(\frac{3 \pi}{2}\right)\text{ we have}\\
    &w_{0}=5 \operatorname{cis}\left(\frac{3 \pi}{4}\right)=-\frac{5 \sqrt{2}}{2}+\frac{5 \sqrt{2}}{2} i \quad\quad\quad\quad\quad w_{1}=5 \operatorname{cis}\left(\frac{7 \pi}{4}\right)=\frac{5 \sqrt{2}}{2}-\frac{5 \sqrt{2}}{2} i
    \end{aligned}\)
67. \(\begin{aligned}
    &\text{Since }z=1+i \sqrt{3}=2 \operatorname{cis}\left(\frac{\pi}{3}\right)\text{ we have}\\
    &w_{0}=\sqrt{2} \operatorname{cis}\left(\frac{\pi}{6}\right)=\frac{\sqrt{6}}{2}+\frac{\sqrt{2}}{2} i \quad\quad\quad\quad\quad w_{1}=\sqrt{2} \operatorname{cis}\left(\frac{7 \pi}{6}\right)=-\frac{\sqrt{6}}{2}-\frac{\sqrt{2}}{2} i
    \end{aligned}\)
68. \(\begin{aligned}
    &\text{Since }z=\frac{5}{2}-\frac{5 \sqrt{3}}{2} i=5 \operatorname{cis}\left(\frac{5 \pi}{3}\right)\text{ we have}\\
    &w_{0}=\sqrt{5} \operatorname{cis}\left(\frac{5 \pi}{6}\right)=-\frac{\sqrt{15}}{2}+\frac{\sqrt{5}}{2} i \quad\quad\quad\quad\quad w_{1}=\sqrt{5} \operatorname{cis}\left(\frac{11 \pi}{6}\right)=\frac{\sqrt{15}}{2}-\frac{\sqrt{5}}{2} i
    \end{aligned}\)
69. \(\begin{aligned}
    &\text{Since }z=64=64 \operatorname{cis}(0)\text{ we have}\\
    &w_{0}=4 \operatorname{cis}(0)=4 \quad\quad\quad\quad w_{1}=4 \operatorname{cis}\left(\frac{2 \pi}{3}\right)=-2+2 i \sqrt{3} \quad\quad\quad\quad w_{2}=4 \operatorname{cis}\left(\frac{4 \pi}{3}\right)=-2-2 i \sqrt{3}
    \end{aligned}\)
70. \(\begin{aligned}
    &\text{Since }z=-125=125 \operatorname{cis}(\pi)\text{ we have}\\
    &w_{0}=5 \operatorname{cis}\left(\frac{\pi}{3}\right)=\frac{5}{2}+\frac{5 \sqrt{3}}{2} i \quad\quad\quad\quad w_{1}=5 \operatorname{cis}(\pi)=-5 \quad\quad\quad\quad w_{2}=5 \operatorname{cis}\left(\frac{5 \pi}{3}\right)=\frac{5}{2}-\frac{5 \sqrt{3}}{2} i
    \end{aligned}\)
71. \(\begin{aligned}
    &\text{Since }z=i=\operatorname{cis}\left(\frac{\pi}{2}\right)\text{ we have}\\
    &w_{0}=\operatorname{cis}\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}+\frac{1}{2} i \quad\quad\quad\quad w_{1}=\operatorname{cis}\left(\frac{5 \pi}{6}\right)=-\frac{\sqrt{3}}{2}+\frac{1}{2} i \quad\quad\quad\quad w_{2}=\operatorname{cis}\left(\frac{3 \pi}{2}\right)=-i
    \end{aligned}\)
72. \(\begin{aligned}
    &\text{Since }z=-8 i=8 \operatorname{cis}\left(\frac{3 \pi}{2}\right)\text{ we have}\\
    &w_{0}=2 \operatorname{cis}\left(\frac{\pi}{2}\right)=2 i \quad\quad\quad\quad w_{1}=2 \operatorname{cis}\left(\frac{7 \pi}{6}\right)=-\sqrt{3}-i \quad\quad\quad\quad w_{2}=\operatorname{cis}\left(\frac{11 \pi}{6}\right)=\sqrt{3}-i
    \end{aligned}\)
73. \(\begin{array}{l}
    &\text{Since }z=16=16 \operatorname{cis}(0)\text{ we have}&\\
    &w_{0}=2 \operatorname{cis}(0)=2 & w_{1}=2 \operatorname{cis}\left(\frac{\pi}{2}\right)=2 i \\
    &w_{2}=2 \operatorname{cis}(\pi)=-2 & w_{3}=2 \operatorname{cis}\left(\frac{3 \pi}{2}\right)=-2 i
    \end{array}\)
74. \(\begin{array}{ll}
    \text{Since }z=-81=81 \operatorname{cis}(\pi)\text{ we have}\\
    w_{0}=3 \operatorname{cis}\left(\frac{\pi}{4}\right)=\frac{3 \sqrt{2}}{2}+\frac{3 \sqrt{2}}{2} i & w_{1}=3 \operatorname{cis}\left(\frac{3 \pi}{4}\right)=-\frac{3 \sqrt{2}}{2}+\frac{3 \sqrt{2}}{2} i \\
    w_{2}=3 \operatorname{cis}\left(\frac{5 \pi}{4}\right)=-\frac{3 \sqrt{2}}{2}-\frac{3 \sqrt{2}}{2} i & w_{3}=3 \operatorname{cis}\left(\frac{7 \pi}{4}\right)=\frac{3 \sqrt{2}}{2}-\frac{3 \sqrt{2}}{2} i
    \end{array}\)
75. \(\begin{array}{ll}
    \text{Since }z=64=64 \operatorname{cis}(0)\text{ we have}\\
    w_{0}=2 \operatorname{cis}(0)=2 & w_{1}=2 \operatorname{cis}\left(\frac{\pi}{3}\right)=1+\sqrt{3} i & w_{2}=2 \operatorname{cis}\left(\frac{2 \pi}{3}\right)=-1+\sqrt{3} i \\
    w_{3}=2 \operatorname{cis}(\pi)=-2 & w_{4}=2 \operatorname{cis}\left(-\frac{2 \pi}{3}\right)=-1-\sqrt{3} i & w_{5}=2 \operatorname{cis}\left(-\frac{\pi}{3}\right)=1-\sqrt{3} i
    \end{array}\)
76. \(\begin{array}{ll}
    \text{Since }z=-729=729 \operatorname{cis}(\pi)\text{ we have}\\
    w_{0}=3 \operatorname{cis}\left(\frac{\pi}{6}\right)=\frac{3 \sqrt{3}}{2}+\frac{3}{2} i & w_{1}=3 \operatorname{cis}\left(\frac{\pi}{2}\right)=3 i & w_{2}=3 \operatorname{cis}\left(\frac{5 \pi}{6}\right)=-\frac{3 \sqrt{3}}{2}+\frac{3}{2} i \\
    w_{3}=3 \operatorname{cis}\left(\frac{7 \pi}{6}\right)=-\frac{3 \sqrt{3}}{2}-\frac{3}{2} i & w_{4}=3 \operatorname{cis}\left(-\frac{3 \pi}{2}\right)=-3 i & w_{5}=3 \operatorname{cis}\left(-\frac{11 \pi}{6}\right)=\frac{3 \sqrt{3}}{2}-\frac{3}{2} i
    \end{array}\)
77. Note: In the answers for \(w_{0}\) and \(w_{2}\) the first rectangular form comes from applying the apporpriate Sum or Difference Identity \(\left(\frac{\pi}{12}=\frac{\pi}{3}-\frac{\pi}{4} \text { and } \frac{17 \pi}{12}=\frac{2 \pi}{3}+\frac{3 \pi}{4}, \text { respectively }\right)\) and the second comes from using the second comes from using the Half-Angle Identities.

    \(w_{0}=\sqrt[3]{2} \operatorname{cis}\left(\frac{\pi}{12}\right)=\sqrt[3]{2}\left(\frac{\sqrt{6}+\sqrt{2}}{4}+i\left(\frac{\sqrt{6}-\sqrt{2}}{4}\right)\right)=\sqrt[3]{2}\left(\frac{\sqrt{2+\sqrt{3}}}{2}+i \frac{\sqrt{2-\sqrt{3}}}{2}\right)\)

    \(w_{1}=\sqrt[3]{2} \operatorname{cis}\left(\frac{3 \pi}{4}\right)=\sqrt[3]{2}\left(-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2} i\right)\)

    \(w_{2}=\sqrt[3]{2} \operatorname{cis}\left(\frac{17 \pi}{12}\right)=\sqrt[3]{2}\left(\frac{\sqrt{2}-\sqrt{6}}{4}+i\left(\frac{-\sqrt{2}-\sqrt{6}}{4}\right)\right)=\sqrt[3]{2}\left(\frac{\sqrt{2-\sqrt{3}}}{2}+i \frac{\sqrt{2+\sqrt{3}}}{2}\right)\)

78. \(\begin{aligned}
    &w_{0}=\operatorname{cis}(0)=1 \\
    &w_{1}=\operatorname{cis}\left(\frac{2 \pi}{5}\right) \approx 0.309+0.951 i \\
    &w_{2}=\operatorname{cis}\left(\frac{4 \pi}{5}\right) \approx-0.809+0.588 i \\
    &w_{3}=\operatorname{cis}\left(\frac{6 \pi}{5}\right) \approx-0.809-0.588 i \\
    &w_{4}=\operatorname{cis}\left(\frac{8 \pi}{5}\right) \approx 0.309-0.951 i
    \end{aligned}\)

79. \(p(x)=x^{12}-4096=(x-2)(x+2)\left(x^{2}+4\right)\left(x^{2}-2 x+4\right)\left(x^{2}+2 x+4\right)\left(x^{2}-2 \sqrt{3} x+4\right)\left(x^{2}+2 \sqrt{3}+4\right)\)