Worksheet: Arbitrary Roots of Complex Numbers

In this worksheet, we will practice using de Moivre’s theorem to find the nth roots of a complex number and exploring their properties.

Q1:

Find the cube roots of 64, giving your answers in trigonometric form.

  • A๐‘=4๏Šง, ๐‘=4๏€ผ๏€ผ4๐œ‹3๏ˆโˆ’๐‘–๏€ผ4๐œ‹3๏ˆ๏ˆ๏Šจsincos, ๐‘=4๏€ผ๏€ผโˆ’4๐œ‹3๏ˆโˆ’๐‘–๏€ผโˆ’4๐œ‹3๏ˆ๏ˆ๏Šฉsincos
  • B๐‘=8๏Šง, ๐‘=8๏€ผ๏€ผ4๐œ‹3๏ˆโˆ’๐‘–๏€ผ4๐œ‹3๏ˆ๏ˆ๏Šจsincos, ๐‘=8๏€ผ๏€ผโˆ’4๐œ‹3๏ˆโˆ’๐‘–๏€ผโˆ’4๐œ‹3๏ˆ๏ˆ๏Šฉsincos
  • C๐‘=8๏Šง, ๐‘=8๏€ผ๏€ผ2๐œ‹3๏ˆ+๐‘–๏€ผ2๐œ‹3๏ˆ๏ˆ๏Šจcossin, ๐‘=8๏€ผ๏€ผโˆ’2๐œ‹3๏ˆ+๐‘–๏€ผโˆ’2๐œ‹3๏ˆ๏ˆ๏Šฉcossin
  • D๐‘=4๏Šง, ๐‘=4๏€ผ๏€ผ2๐œ‹3๏ˆ+๐‘–๏€ผ2๐œ‹3๏ˆ๏ˆ๏Šจcossin, ๐‘=4๏€ผ๏€ผโˆ’2๐œ‹3๏ˆ+๐‘–๏€ผโˆ’2๐œ‹3๏ˆ๏ˆ๏Šฉcossin

Q2:

Find the fourth roots of โˆ’1, giving your answers in trigonometric form.

  • A๐‘=๏€ผ๏€ผ3๐œ‹4๏ˆโˆ’๐‘–๏€ผ3๐œ‹4๏ˆ๏ˆ๏Šงsincos, ๐‘=๏€ผ๏€ผ5๐œ‹4๏ˆโˆ’๐‘–๏€ผ5๐œ‹4๏ˆ๏ˆ๏Šจsincos, ๐‘=๏€ผ๏€ผ7๐œ‹4๏ˆโˆ’๐‘–๏€ผ7๐œ‹4๏ˆ๏ˆ๏Šฉsincos, ๐‘=๏€ผ๏€ผ9๐œ‹4๏ˆโˆ’๐‘–๏€ผ9๐œ‹4๏ˆ๏ˆ๏Šชsincos
  • B๐‘=2๏€ผ๏€ผ3๐œ‹4๏ˆโˆ’๐‘–๏€ผ3๐œ‹4๏ˆ๏ˆ๏Šงsincos, ๐‘=2๏€ผ๏€ผ5๐œ‹4๏ˆโˆ’๐‘–๏€ผ5๐œ‹4๏ˆ๏ˆ๏Šจsincos, ๐‘=2๏€ผ๏€ผ7๐œ‹4๏ˆโˆ’๐‘–๏€ผ7๐œ‹4๏ˆ๏ˆ๏Šฉsincos, ๐‘=2๏€ผ๏€ผ9๐œ‹4๏ˆโˆ’๐‘–๏€ผ9๐œ‹4๏ˆ๏ˆ๏Šชsincos
  • C๐‘=2๏€ป๏€ป๐œ‹4๏‡+๐‘–๏€ป๐œ‹4๏‡๏‡๏Šงcossin, ๐‘=2๏€ผ๏€ผ3๐œ‹4๏ˆ+๐‘–๏€ผ3๐œ‹4๏ˆ๏ˆ๏Šจcossin, ๐‘=2๏€ป๏€ปโˆ’๐œ‹4๏‡+๐‘–๏€ปโˆ’๐œ‹4๏‡๏‡๏Šฉcossin, ๐‘=2๏€ผ๏€ผโˆ’3๐œ‹4๏ˆ+๐‘–๏€ผโˆ’3๐œ‹4๏ˆ๏ˆ๏Šชcossin
  • D๐‘=๏€ป๏€ป๐œ‹4๏‡+๐‘–๏€ป๐œ‹4๏‡๏‡๏Šงcossin, ๐‘=๏€ผ๏€ผ3๐œ‹4๏ˆ+๐‘–๏€ผ3๐œ‹4๏ˆ๏ˆ๏Šจcossin, ๐‘=๏€ป๏€ปโˆ’๐œ‹4๏‡+๐‘–๏€ปโˆ’๐œ‹4๏‡๏‡๏Šฉcossin, ๐‘=๏€ผ๏€ผโˆ’3๐œ‹4๏ˆ+๐‘–๏€ผโˆ’3๐œ‹4๏ˆ๏ˆ๏Šชcossin

Q3:

Determine the solution set of the equation ๐‘ง=4๏€ปโˆš2โˆ’โˆš2๐‘–๏‡๏Šฉ in โ„‚, expressing the solutions in exponential form.

  • A๏ญ2๐‘’,2๐‘’,2๐‘’๏น๏Šฑ๏ƒ๏ƒ๏Šฑ๏ƒ๏‘ฝ๏Ž ๏Žก๏Žฆ๏‘ฝ๏Ž ๏Žก๏Žข๏‘ฝ๏Žฃ
  • B๏ญ2๐‘’,2๐‘’,2๐‘’๏น๏Šฑ๏ƒ๏ƒ๏ƒ๏Žฆ๏‘ฝ๏Ž ๏Žก๏‘ฝ๏Ž ๏Žก๏Žข๏‘ฝ๏Žฃ
  • C๏ญ๐‘’,๐‘’,๐‘’๏น๏Šฑ๏ƒ๏ƒ๏ƒ๏‘ฝ๏Ž ๏Žก๏‘ฝ๏Žฃ๏Žฆ๏‘ฝ๏Ž ๏Žก
  • D๏ญ๐‘’,๐‘’,๐‘’๏น๏Šฑ๏ƒ๏ƒ๏Šฑ๏ƒ๏‘ฝ๏Ž ๏Žก๏Žฆ๏‘ฝ๏Ž ๏Žก๏Žข๏‘ฝ๏Žฃ

Q4:

Determine the solution set of the equation ๐‘ง=2+2โˆš3๐‘–๏Šจ in โ„‚.

  • A๏ซโˆš3โˆ’๐‘–,โˆ’โˆš3+๐‘–๏ท
  • B๏ฏโˆš32+12๐‘–,โˆ’โˆš32โˆ’12๐‘–๏ป
  • C๏ฏโˆ’โˆš32+12๐‘–,โˆš32โˆ’12๐‘–๏ป
  • D๏ซโˆš3+๐‘–,โˆ’โˆš3โˆ’๐‘–๏ท

Q5:

Determine the square roots of ๐‘ง, given that ๐‘ง=โˆ’8๐‘–.

  • A{1,โˆ’1}
  • B{๐‘–,โˆ’๐‘–}
  • C๏ฏโˆš32+12๐‘–,โˆ’โˆš32โˆ’12๐‘–๏ป
  • D{2โˆ’2๐‘–,โˆ’2+2๐‘–}
  • E๏ฏ1โˆš2โˆ’1โˆš2๐‘–,โˆ’1โˆš2+1โˆš2๐‘–๏ป

Q6:

Given that ๐‘ง=โˆ’8๐‘–, determine the square roots of ๐‘ง without first converting it to trigonometric form.

  • A{1,โˆ’1}
  • B๏ฏโˆ’12+โˆš32๐‘–,12โˆ’โˆš32๐‘–๏ป
  • C{๐‘–,โˆ’๐‘–}
  • D๏ฏ1โˆš2โˆ’1โˆš2๐‘–,โˆ’1โˆš2+1โˆš2๐‘–๏ป
  • E{2โˆ’2๐‘–,โˆ’2+2๐‘–}

Q7:

Given that ๐‘ง=โˆ’28+96๐‘–, determine the square roots of ๐‘ง without first converting it to trigonometric form.

  • A๏€ผโˆ’725+2425๐‘–๏ˆ,โˆ’๏€ผโˆ’725+2425๐‘–๏ˆ
  • B(6+8๐‘–),โˆ’(6+8๐‘–)
  • C(8+6๐‘–),โˆ’(8+6๐‘–)
  • D๏€ผ2425โˆ’725๐‘–๏ˆ,โˆ’๏€ผ2425โˆ’725๐‘–๏ˆ
  • E๏€ผโˆš2โˆ’12๐‘–๏ˆ,โˆ’๏€ผโˆš2โˆ’12๐‘–๏ˆ

Q8:

Without first converting ๐‘ง to trigonometric form, determine the square roots of ๐‘ง, where ๐‘ง=โˆ’8+2๐‘–1+4๐‘–.

  • A๐‘–, โˆ’๐‘–
  • B1โˆ’๐‘–, 1+๐‘–
  • C1, โˆ’1
  • D1+๐‘–, โˆ’1โˆ’๐‘–

Q9:

Determine the two square roots of 2(โˆ’7โˆ’7๐‘–)โˆ’7+7๐‘– without first converting it to trigonometric form.

  • Aยฑ(1โˆ’๐‘–)
  • Bยฑ(1+๐‘–)
  • Cยฑ1
  • Dยฑ๏€ฟ1โˆš2โˆ’โˆš3๐‘–๏‹
  • Eยฑ๐‘–

Q10:

Find the square roots of โˆ’181โˆ’โˆš3๐‘–, giving your answers in trigonometric form.

  • A3(30+๐‘–30)cossinโˆ˜โˆ˜, 3(210+๐‘–210)cossinโˆ˜โˆ˜
  • B3(60+๐‘–60)cossinโˆ˜โˆ˜, 3(210+๐‘–210)cossinโˆ˜โˆ˜
  • C9(30+๐‘–30)cossinโˆ˜โˆ˜, 9(210+๐‘–210)cossinโˆ˜โˆ˜
  • D(30+๐‘–30)cossinโˆ˜โˆ˜, (210+๐‘–210)cossinโˆ˜โˆ˜
  • E3(30+๐‘–30)cossinโˆ˜โˆ˜, 3(150+๐‘–150)cossinโˆ˜โˆ˜

Q11:

Determine, in trigonometric form, the square roots of 4๐‘’๏Žค๏‘ฝ๏Žฅ๏ƒ.

  • A2๏€ผ๏€ผ5๐œ‹12๏ˆ+๐‘–๏€ผ5๐œ‹12๏ˆ๏ˆcossin, 2๏€ผ๏€ผโˆ’7๐œ‹12๏ˆ+๐‘–๏€ผโˆ’7๐œ‹12๏ˆ๏ˆcossin
  • B๏€ผ๏€ผ5๐œ‹12๏ˆ+๐‘–๏€ผ5๐œ‹12๏ˆ๏ˆcossin, ๏€ผ๏€ผโˆ’7๐œ‹12๏ˆ+๐‘–๏€ผโˆ’7๐œ‹12๏ˆ๏ˆcossin
  • C2๏€ผ๏€ผ7๐œ‹12๏ˆ+๐‘–๏€ผ7๐œ‹12๏ˆ๏ˆcossin, 2๏€ผ๏€ผโˆ’5๐œ‹12๏ˆ+๐‘–๏€ผโˆ’5๐œ‹12๏ˆ๏ˆcossin
  • D๏€ผ๏€ผ5๐œ‹12๏ˆ+๐‘–๏€ผ5๐œ‹12๏ˆ๏ˆcossin, ๏€ผ๏€ผ11๐œ‹12๏ˆ+๐‘–๏€ผ11๐œ‹12๏ˆ๏ˆcossin
  • E2๏€ผ๏€ผ5๐œ‹12๏ˆ+๐‘–๏€ผ5๐œ‹12๏ˆ๏ˆcossin, 2๏€ผ๏€ผ11๐œ‹12๏ˆ+๐‘–๏€ผ11๐œ‹12๏ˆ๏ˆcossin

Q12:

Use De Moivreโ€™s theorem to find the two square roots of 16๏€ผ5๐œ‹3โˆ’๐‘–5๐œ‹3๏ˆcossin.

  • Aยฑ4๐‘–
  • Bยฑ๏€ฟ12โˆ’1โˆš2๐‘–๏‹
  • Cยฑ๏€ป2โˆš3+2๐‘–๏‡
  • Dยฑ๏€ฟโˆš32+12๐‘–๏‹

Q13:

Given that ๐‘ฅ=7โˆ’9๐‘–4โˆ’2๐‘–โˆ’โˆ’8โˆ’5๐‘–โˆ’1+3๐‘–, find the two square roots of ๐‘ฅ without first expressing ๐‘ฅ in trigonometric form.

  • Aยฑ(2โˆ’๐‘–)
  • Bยฑ๏€ฟโˆš32โˆ’1โˆš2๐‘–๏‹
  • Cยฑ๏€ผโˆ’45+35๐‘–๏ˆ
  • Dยฑ๏€ผ35โˆ’45๐‘–๏ˆ
  • Eยฑ(โˆ’1โˆ’2๐‘–)

Q14:

Given that ๐‘ฅ=3+4๐‘–, find ๐‘ฅ๏Šฑ๏Ž ๏Žก.

  • Aยฑ๏€ฝ2โˆ’๐‘–5๏‰
  • Bยฑ๏€ผ45+35๐‘–๏ˆ
  • Cยฑ๏€ฝ1+2๐‘–3๏‰
  • Dยฑ(2+๐‘–)
  • Eยฑ๏€ฟโˆ’1โˆš2+1โˆš2๐‘–๏‹

Q15:

Given that ๐‘ฅ+๐‘ฆ๐‘–=๏€ฝ4+2๐‘–1โˆ’2๐‘–๏‰๏Ž ๏Žก, find all possible real values of ๐‘ฅ and ๐‘ฆ.

  • A{(โˆ’3,โˆ’4),(3,4)}
  • B๏ฌ๏€ผโˆ’โˆš3,โˆ’12๏ˆ,๏€ผโˆš3,12๏ˆ๏ธ
  • C๏ฌ๏€ผ14,โˆ’1๏ˆ,๏€ผโˆ’14,1๏ˆ๏ธ
  • D{(0,2),(0,โˆ’2)}
  • E{(1,1),(โˆ’1,โˆ’1)}

Q16:

Given that ๐‘ง=๏€ป๏€ป๐œ‹6๏‡+๐‘–๏€ป๐œ‹6๏‡๏‡sincos, determine the cubic roots of ๏€น๐‘ง๏…๏Šซ.

  • A๏€ผ๏€ผ5๐œ‹9๏ˆ+๐‘–๏€ผ5๐œ‹9๏ˆ๏ˆcossin, ๏€ผ๏€ผโˆ’7๐œ‹9๏ˆ+๐‘–๏€ผโˆ’7๐œ‹9๏ˆ๏ˆcossin, ๏€ป๏€ปโˆ’๐œ‹9๏‡+๐‘–๏€ปโˆ’๐œ‹9๏‡๏‡cossin
  • B๏€ผ๏€ผ7๐œ‹9๏ˆ+๐‘–๏€ผ7๐œ‹9๏ˆ๏ˆcossin, ๏€ผ๏€ผโˆ’5๐œ‹9๏ˆ+๐‘–๏€ผโˆ’5๐œ‹9๏ˆ๏ˆcossin, ๏€ป๏€ป๐œ‹9๏‡+๐‘–๏€ป๐œ‹9๏‡๏‡cossin
  • C๏€ผ๏€ผ5๐œ‹18๏ˆ+๐‘–๏€ผ5๐œ‹18๏ˆ๏ˆcossin, ๏€ผ๏€ผ17๐œ‹18๏ˆ+๐‘–๏€ผ17๐œ‹18๏ˆ๏ˆcossin, ๏€ผ๏€ผโˆ’7๐œ‹18๏ˆ+๐‘–๏€ผโˆ’7๐œ‹18๏ˆ๏ˆcossin
  • D๏€ผ๏€ผ7๐œ‹9๏ˆ+๐‘–๏€ผ7๐œ‹9๏ˆ๏ˆcossin, ๏€ผ๏€ผโˆ’8๐œ‹9๏ˆ+๐‘–๏€ผโˆ’8๐œ‹9๏ˆ๏ˆcossin, ๏€ผ๏€ผโˆ’5๐œ‹9๏ˆ+๐‘–๏€ผโˆ’5๐œ‹9๏ˆ๏ˆcossin
  • E๏€ผ๏€ผโˆ’7๐œ‹9๏ˆ+๐‘–๏€ผโˆ’7๐œ‹9๏ˆ๏ˆcossin, ๏€ผ๏€ผโˆ’4๐œ‹9๏ˆ+๐‘–๏€ผโˆ’4๐œ‹9๏ˆ๏ˆcossin, ๏€ป๏€ปโˆ’๐œ‹9๏‡+๐‘–๏€ปโˆ’๐œ‹9๏‡๏‡cossin

Q17:

Given that ๐‘ง=โˆ’5+4๐‘–โˆ’5๐‘–2+7๐‘–โˆ’2๐‘–+7๐‘–๏Šจ๏Šจ๏Šฉ, find the square roots of ๐‘ง in trigonometric form.

  • A๏€ป๏€ป๐œ‹4๏‡+๐‘–๏€ป๐œ‹4๏‡๏‡cossin, ๏€ผ๏€ผโˆ’3๐œ‹4๏ˆ+๐‘–๏€ผโˆ’3๐œ‹4๏ˆ๏ˆcossin
  • B๏€ป๏€ป๐œ‹2๏‡+๐‘–๏€ป๐œ‹2๏‡๏‡cossin, ๏€ป๏€ปโˆ’๐œ‹2๏‡+๐‘–๏€ปโˆ’๐œ‹2๏‡๏‡cossin
  • C๏€ป๏€ป๐œ‹3๏‡+๐‘–๏€ป๐œ‹3๏‡๏‡cossin, ๏€ป๏€ปโˆ’๐œ‹3๏‡+๐‘–๏€ปโˆ’๐œ‹3๏‡๏‡cossin
  • D๏€ป๏€ป๐œ‹6๏‡+๐‘–๏€ป๐œ‹6๏‡๏‡cossin, ๏€ป๏€ปโˆ’๐œ‹6๏‡+๐‘–๏€ปโˆ’๐œ‹6๏‡๏‡cossin
  • E๏€ป๏€ปโˆ’๐œ‹4๏‡+๐‘–๏€ปโˆ’๐œ‹4๏‡๏‡cossin, ๏€ผ๏€ผ3๐œ‹4๏ˆ+๐‘–๏€ผ3๐œ‹4๏ˆ๏ˆcossin

Q18:

Find the two square roots of ๐‘ฅ=8+๐‘–โˆ’1โˆ’2๐‘–+โˆ’9+2๐‘–โˆ’4โˆ’๐‘– in โ„‚, without first expressing ๐‘ฅ in trigonometric form.

  • Aยฑ(1+๐‘–)
  • Bยฑ(1โˆ’๐‘–)
  • Cยฑ๐‘–
  • Dยฑ1

Q19:

Find the possible values of 1โˆš3๏€ฝ(๐‘–)+(๐‘–)๏‰๏Ž ๏Žข๏Ž ๏Žข๏Šฑ.

  • Aโˆ’2, 0, and 2
  • Bโˆ’1, 0, and 1
  • Cโˆ’2โˆš3, 0, and 2โˆš3
  • Dโˆ’1โˆš3, 0, and 1โˆš3
  • Eโˆ’13, 0, and 13

Q20:

Find the solutions to the equation ๐‘ง=125๐‘’๏Šฌ๏ƒ๏Žก๏‘ฝ๏Žข. What are their geometrical properties?

  • A5๐‘’๏‘ฝ๏Žจ๏ƒ, 5๐‘’๏Žฃ๏‘ฝ๏Žจ๏ƒ, 5๐‘’๏Žฆ๏‘ฝ๏Žจ๏ƒ, 5๐‘’๏Šฑ๏ƒ๏Žก๏‘ฝ๏Žจ, 5๐‘’๏Šฑ๏ƒ๏Žค๏‘ฝ๏Žจ, 5๐‘’๏Šฑ๏ƒ๏Žง๏‘ฝ๏Žจ; the roots lie at the vertices of a regular hexagon centered at the origin, inscribed in a circle of radius 5.
  • Bโˆš5๐‘’๏‘ฝ๏Žจ๏ƒ, โˆš5๐‘’๏Žฃ๏‘ฝ๏Žจ๏ƒ, โˆš5๐‘’๏Žฆ๏‘ฝ๏Žจ๏ƒ, โˆš5๐‘’๏Šฑ๏ƒ๏Žก๏‘ฝ๏Žจ, โˆš5๐‘’๏Šฑ๏ƒ๏Žค๏‘ฝ๏Žจ, โˆš5๐‘’๏Šฑ๏ƒ๏Žง๏‘ฝ๏Žจ; the roots lie at the vertices of a regular hexagon centered at the origin, inscribed in a circle of radius โˆš5.
  • C5๐‘’๏‘ฝ๏Žจ๏ƒ, 5๐‘’๏Žฃ๏‘ฝ๏Žจ๏ƒ, 5๐‘’๏Žฆ๏‘ฝ๏Žจ๏ƒ, 5๐‘’๏Šฑ๏ƒ๏Žก๏‘ฝ๏Žจ, 5๐‘’๏Šฑ๏ƒ๏Žค๏‘ฝ๏Žจ, 5๐‘’๏Šฑ๏ƒ๏Žง๏‘ฝ๏Žจ; the roots lie at the vertices of a regular hexagon centered at the origin, inscribed in a circle of radius โˆš5.
  • Dโˆš5๐‘’๏‘ฝ๏Žจ๏ƒ, โˆš5๐‘’๏Žค๏‘ฝ๏Ž ๏Žง๏ƒ, โˆš5๐‘’๏Žฃ๏‘ฝ๏Žจ๏ƒ, โˆš5๐‘’๏Ž ๏Ž ๏‘ฝ๏Ž ๏Žง๏ƒ, โˆš5๐‘’๏Žฆ๏‘ฝ๏Žจ๏ƒ, โˆš5๐‘’๏Ž ๏Žฆ๏‘ฝ๏Ž ๏Žง๏ƒ; the roots lie at the vertices of a regular hexagon centered at the origin, inscribed in a circle of radius โˆš5.
  • E125๐‘’๏Žก๏‘ฝ๏Žข๏ƒ, 125๐‘’๏Žง๏‘ฝ๏Žข๏ƒ, 125๐‘’๏Žฆ๏‘ฝ๏Žจ๏ƒ, 125๐‘’๏Šฑ๏ƒ๏Ž ๏Žฃ๏‘ฝ๏Žข, 125๐‘’๏Šฑ๏ƒ๏Žค๏‘ฝ๏Žข, 125๐‘’๏Šฑ๏ƒ๏Žง๏‘ฝ๏Žจ; the roots lie at a straight line that passes through the origin.

State the 6th roots of unity.

  • A๐‘’๏‘ฝ๏Žฅ๏ƒ, ๐‘–, ๐‘’๏Žค๏‘ฝ๏Žฅ๏ƒ, ๐‘’๏Šฑ๏ƒ๏‘ฝ๏Žฅ, โˆ’๐‘–, ๐‘’๏Šฑ๏ƒ๏Žค๏‘ฝ๏Žฅ
  • B1, ๐‘’๏‘ฝ๏Žข๏ƒ, ๐‘–, โˆ’1, ๐‘’๏Šฑ๏ƒ๏‘ฝ๏Žข, โˆ’๐‘–
  • C1, ๐‘’๏‘ฝ๏Žฅ๏ƒ, ๐‘’๏‘ฝ๏Žข๏ƒ, ๐‘’๏‘ฝ๏Žก๏ƒ, ๐‘’๏Žก๏‘ฝ๏Žข๏ƒ, ๐‘’๏Žค๏‘ฝ๏Žฅ๏ƒ
  • D1, ๐‘’๏‘ฝ๏Žฅ๏ƒ, ๐‘’๏Žค๏‘ฝ๏Žฅ๏ƒ, โˆ’1, ๐‘’๏Šฑ๏ƒ๏‘ฝ๏Žฅ, ๐‘’๏Šฑ๏ƒ๏Žค๏‘ฝ๏Žฅ
  • E1, ๐‘’๏‘ฝ๏Žข๏ƒ, ๐‘’๏Žก๏‘ฝ๏Žข๏ƒ, โˆ’1, ๐‘’๏Šฑ๏ƒ๏‘ฝ๏Žข, ๐‘’๏Šฑ๏ƒ๏Žก๏‘ฝ๏Žข

What is the relationship between the 6th roots of unity and the solutions to the equation ๐‘ง=125๐‘’๏Šฌ๏ƒ๏Žก๏‘ฝ๏Žข?

  • AThe solutions to the equation are the 6th roots of unity multiplied by 5๐‘’๏‘ฝ๏Žจ๏ƒ.
  • BThe solutions to the equation are the 6th roots of unity multiplied by 125๐‘’๏Žก๏‘ฝ๏Žข๏ƒ.
  • CThe solutions to the equation are the 6th roots of unity multiplied by 5๐‘’๏‘ฝ๏Žข.
  • DThe solutions to the equation are the 6th roots of unity multiplied by โˆš5๐‘’๏‘ฝ๏Žจ.
  • EThe solutions to the equation are the 6th roots of unity multiplied by โˆš5๐‘’๏‘ฝ๏Žจ๏ƒ.

Q21:

Find the roots of ๐‘ง+16=0๏Šฎ.

  • Aโˆš2๐‘’๏‘ฝ๏Žง๏ƒ, โˆš2๐‘’๏Žข๏‘ฝ๏Žง๏ƒ, โˆš2๐‘’๏Žค๏‘ฝ๏Žง๏ƒ, โˆš2๐‘’๏Žฆ๏‘ฝ๏Žง๏ƒ, โˆš2๐‘’๏Šฑ๏ƒ๏Žฆ๏‘ฝ๏Žง, โˆš2๐‘’๏Šฑ๏ƒ๏Žค๏‘ฝ๏Žง, โˆš2๐‘’๏Šฑ๏ƒ๏Žข๏‘ฝ๏Žง, โˆš2๐‘’๏Šฑ๏ƒ๏‘ฝ๏Žง.
  • Bโˆš2๐‘’๏‘ฝ๏Žฃ๏ƒ, โˆš2๐‘’๏Žข๏‘ฝ๏Žฃ๏ƒ, โˆš2๐‘’๏Žค๏‘ฝ๏Žฃ๏ƒ, โˆš2๐‘’๏Žฆ๏‘ฝ๏Žฃ๏ƒ, โˆš2๐‘’๏Šฑ๏ƒ๏Žฆ๏‘ฝ๏Žฃ, โˆš2๐‘’๏Šฑ๏ƒ๏Žค๏‘ฝ๏Žฃ, โˆš2๐‘’๏Šฑ๏ƒ๏Žข๏‘ฝ๏Žฃ, โˆš2๐‘’๏Šฑ๏ƒ๏‘ฝ๏Žฃ
  • C2๐‘’๏‘ฝ๏Žง๏ƒ, 2๐‘’๏Žข๏‘ฝ๏Žง๏ƒ, 2๐‘’๏Žค๏‘ฝ๏Žง๏ƒ, 2๐‘’๏Žฆ๏‘ฝ๏Žง๏ƒ, 2๐‘’๏Šฑ๏ƒ๏Žฆ๏‘ฝ๏Žง, 2๐‘’๏Šฑ๏ƒ๏Žค๏‘ฝ๏Žง, 2๐‘’๏Šฑ๏ƒ๏Žข๏‘ฝ๏Žง, 2๐‘’๏Šฑ๏ƒ๏‘ฝ๏Žง
  • D๐‘’๏‘ฝ๏Ž ๏Žฅ, ๐‘’๏Žข๏‘ฝ๏Ž ๏Žฅ, ๐‘’๏Žค๏‘ฝ๏Ž ๏Žฅ, ๐‘’๏Žฆ๏‘ฝ๏Ž ๏Žฅ, ๐‘’๏Žจ๏‘ฝ๏Ž ๏Žฅ, ๐‘’๏Ž ๏Ž ๏‘ฝ๏Ž ๏Žฅ, ๐‘’๏Ž ๏Žข๏‘ฝ๏Ž ๏Žฅ, ๐‘’๏Ž ๏Žค๏‘ฝ๏Ž ๏Žฅ
  • E2๐‘’๏‘ฝ๏Žฃ๏ƒ, 2๐‘’๏Žข๏‘ฝ๏Žฃ๏ƒ, 2๐‘’๏Žค๏‘ฝ๏Žฃ๏ƒ, 2๐‘’๏Žฆ๏‘ฝ๏Žฃ๏ƒ, 2๐‘’๏Šฑ๏ƒ๏Žฆ๏‘ฝ๏Žฃ, 2๐‘’๏Šฑ๏ƒ๏Žค๏‘ฝ๏Žฃ, 2๐‘’๏Šฑ๏ƒ๏Žข๏‘ฝ๏Žฃ, 2๐‘’๏Šฑ๏ƒ๏‘ฝ๏Žฃ

The complex numbers representing the roots of ๐‘ง+16=0๏Šฎ are each squared to form the vertices of a new shape. What is the area of this shape?

Q22:

Find the coordinates of the vertices of a regular hexagon centered at (โˆ’1,2) with one vertex at the origin.

Give your answer as exact Cartesian coordinates.

  • A(0,0), ๏€ฟ3+2โˆš32,โˆš3โˆ’62๏‹, ๏€ฟ1+2โˆš32,โˆš3โˆ’22๏‹, (2,โˆ’4), ๏€ฟ1โˆ’2โˆš32,โˆ’โˆš3โˆ’22๏‹, ๏€ฟ3โˆ’2โˆš32,โˆ’โˆš3โˆ’62๏‹
  • B(0,0), ๏€ฟโˆ’2โˆ’โˆš32,โˆ’2โˆš3+12๏‹, ๏€ฟโˆ’6โˆ’โˆš32,โˆ’2โˆš3+12๏‹, (โˆ’4,2), ๏€ฟโˆ’6+โˆš32,2โˆš3+32๏‹, ๏€ฟโˆ’2+โˆš32,2โˆš3+12๏‹
  • C(0,0), ๏€ฟโˆ’1+2โˆš32,โˆš3+22๏‹, ๏€ฟโˆ’3+2โˆš32,โˆš3+62๏‹, (โˆ’2,4), ๏€ฟโˆ’3โˆ’2โˆš32,โˆ’โˆš3+62๏‹, ๏€ฟโˆ’1โˆ’2โˆš32,โˆ’โˆš3+22๏‹
  • D(0,0), (โˆ’2,0), (โˆ’3,2), (โˆ’2,4), (0,4), (1,2)
  • E(0,0), ๏€ฟ2+โˆš32,2โˆš3โˆ’12๏‹, ๏€ฟ6+โˆš32,2โˆš3โˆ’32๏‹, (4,โˆ’2), ๏€ฟ6โˆ’โˆš32,โˆ’2โˆš3โˆ’32๏‹, ๏€ฟ2โˆ’โˆš32,โˆ’2โˆš3โˆ’12๏‹

Q23:

Solve ๐‘ง=16โˆš2+16๐‘–โˆš2๏Šซ.

  • A๐‘ง=2๐‘’๏‘ฝ๏Žก๏ŽŸ๏ƒ, 2๐‘’๏Žจ๏‘ฝ๏Žก๏ŽŸ๏ƒ, 2๐‘’๏Ž ๏Žฆ๏‘ฝ๏Žก๏ŽŸ๏ƒ, 2๐‘’๏Šฑ๏ƒ๏Žฆ๏‘ฝ๏Žก๏ŽŸ, 2๐‘’๏Šฑ๏ƒ๏Žข๏‘ฝ๏Žฃ
  • B๐‘ง=2๐‘’๏‘ฝ๏Ž ๏Žค๏ƒ, 2๐‘’๏Žฆ๏‘ฝ๏Ž ๏Žค๏ƒ, 2๐‘’๏Ž ๏Žข๏‘ฝ๏Ž ๏Žค๏ƒ, 2๐‘’๏Šฑ๏ƒ๏Ž ๏‘ฝ๏Žข, 2๐‘’๏Šฑ๏ƒ๏Ž ๏Ž ๏‘ฝ๏Ž ๏Žค
  • C๐‘ง=4๐‘’๏‘ฝ๏Žก๏ŽŸ๏ƒ, 4๐‘’๏Žจ๏‘ฝ๏Žก๏ŽŸ๏ƒ, 4๐‘’๏Ž ๏Žฆ๏‘ฝ๏Žก๏ŽŸ๏ƒ, 4๐‘’๏Šฑ๏ƒ๏Žฆ๏‘ฝ๏Žก๏ŽŸ, 4๐‘’๏Šฑ๏ƒ๏Žข๏‘ฝ๏Žฃ
  • D๐‘ง=4๐‘’๏‘ฝ๏Ž ๏Žค๏ƒ, 4๐‘’๏Žฆ๏‘ฝ๏Ž ๏Žค๏ƒ, 4๐‘’๏Ž ๏Žข๏‘ฝ๏Ž ๏Žค๏ƒ, 4๐‘’๏Šฑ๏ƒ๏Ž ๏‘ฝ๏Žข, 4๐‘’๏Šฑ๏ƒ๏Ž ๏Ž ๏‘ฝ๏Ž ๏Žค
  • E๐‘ง=32๐‘’๏‘ฝ๏Žฃ๏ƒ, 32๐‘’๏Žจ๏‘ฝ๏Žฃ๏ƒ, 32๐‘’๏Ž ๏Žฆ๏‘ฝ๏Žฃ๏ƒ, 32๐‘’๏Šฑ๏ƒ๏Žฆ๏‘ฝ๏Žฃ, 32๐‘’๏Šฑ๏Šฉ๏Ž„๏ƒ

By plotting these solutions on an Argand diagram, or otherwise, describe the geometric properties of the solutions.

  • AThe roots lie at the vertices of a regular pentagon inscribed in a circle of radius 32 at the origin.
  • BThe roots lie at the vertices of a regular pentagon inscribed in a circle of radius 1 at the origin.
  • CThe roots lie at the vertices of a regular pentagon inscribed in a circle of radius 4 at the origin.
  • DThe roots lie in a straight line.
  • EThe roots lie at the vertices of a regular pentagon inscribed in a circle of radius 2 at the origin.

Q24:

A small robot is programmed to travel ๐‘Ž units forward and then turn to the left by an angle of 2๐œ‘. If it does this ๐‘› times, how far will it be from the starting point? Give you answer in exact form.

  • A|||(๐‘›๐œ‘)๐œ‘|||sinsin
  • B|(๐‘›๐œ‘)|cos
  • C|๐‘Ž||||(๐‘›๐œ‘)๐œ‘|||sinsin
  • D|๐‘Ž|
  • E|๐‘Ž||(๐‘›๐œ‘)|sin

Q25:

Find the coordinates of the vertices of a regular pentagon centered at the origin with one vertex at (3,3).

Give your answer as exact Cartesian coordinates.

  • A(3,3), ๏€ผ3โˆš2๏€ผ13๐œ‹20๏ˆ,3โˆš2๏€ผ13๐œ‹20๏ˆ๏ˆsincos, ๏€ผ3โˆš2๏€ผโˆ’19๐œ‹20๏ˆ,3โˆš2๏€ผโˆ’19๐œ‹20๏ˆ๏ˆsincos, ๏€ผ3โˆš2๏€ผโˆ’11๐œ‹20๏ˆ,3โˆš2๏€ผโˆ’11๐œ‹20๏ˆ๏ˆsincos, ๏€ผ3โˆš2๏€ผโˆ’3๐œ‹20๏ˆ,3โˆš2๏€ผโˆ’3๐œ‹20๏ˆ๏ˆsincos
  • B(3,3), ๏€ผ3โˆš2๏€ผ13๐œ‹20๏ˆ,3โˆš2๏€ผ13๐œ‹20๏ˆ๏ˆcossin, ๏€ผ3โˆš2๏€ผโˆ’19๐œ‹20๏ˆ,3โˆš2๏€ผโˆ’19๐œ‹20๏ˆ๏ˆcossin, ๏€ผ3โˆš2๏€ผโˆ’11๐œ‹20๏ˆ,3โˆš2๏€ผโˆ’11๐œ‹20๏ˆ๏ˆcossin, ๏€ผ3โˆš2๏€ผโˆ’3๐œ‹20๏ˆ,3โˆš2๏€ผโˆ’3๐œ‹20๏ˆ๏ˆcossin
  • C(3,3), ๏€ผ3โˆš2๏€ผ11๐œ‹15๏ˆ,3โˆš2๏€ผ11๐œ‹15๏ˆ๏ˆcossin, ๏€ผ3โˆš2๏€ผโˆ’13๐œ‹15๏ˆ,3โˆš2๏€ผโˆ’13๐œ‹15๏ˆ๏ˆcossin, ๏€ผ3โˆš2๏€ผโˆ’7๐œ‹15๏ˆ,3โˆš2๏€ผโˆ’7๐œ‹15๏ˆ๏ˆcossin, ๏€ป3โˆš2๏€ปโˆ’๐œ‹15๏‡,3โˆš2๏€ปโˆ’๐œ‹15๏‡๏‡cossin
  • D(3,3), ๏€ผ3๏€ผ13๐œ‹20๏ˆ,3๏€ผ13๐œ‹20๏ˆ๏ˆsincos, ๏€ผ3๏€ผโˆ’19๐œ‹20๏ˆ,3๏€ผโˆ’19๐œ‹20๏ˆ๏ˆsincos, ๏€ผ3๏€ผโˆ’11๐œ‹20๏ˆ,3๐‘๐‘œ๐‘ ๏€ผโˆ’11๐œ‹20๏ˆ๏ˆsin, ๏€ผ3๏€ผโˆ’3๐œ‹20๏ˆ,3๏€ผโˆ’3๐œ‹20๏ˆ๏ˆsincos
  • E(3,3), ๏€ผ3๏€ผ13๐œ‹20๏ˆ,3๏€ผ13๐œ‹20๏ˆ๏ˆcossin, ๏€ผ3๏€ผโˆ’19๐œ‹20๏ˆ,3๏€ผโˆ’19๐œ‹20๏ˆ๏ˆcossin, ๏€ผ3๏€ผโˆ’11๐œ‹20๏ˆ,3๏€ผโˆ’11๐œ‹20๏ˆ๏ˆcossin, ๏€ผ3๏€ผโˆ’3๐œ‹20๏ˆ,3๏€ผโˆ’3๐œ‹20๏ˆ๏ˆcossin

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