Lesson Worksheet: De Moivre’s Theorem Mathematics

In this worksheet, we will practice finding powers and roots of complex numbers using De Moivre’s theorem to simplify calculations of powers and roots.

Q1:

Use De Moivreโ€™s theorem to find the two square roots of 9๏€ผ2๐œ‹3+๐‘–2๐œ‹3๏ˆcossin.

  • A๏ฏ32+3โˆš32๐‘–,โˆ’32+3โˆš32๐‘–๏ป
  • B๏ฏโˆš32โˆ’12๐‘–,โˆ’โˆš32+12๐‘–๏ป
  • C{โˆ’3,3}
  • D๏ฏ32+3โˆš32๐‘–,โˆ’32โˆ’3โˆš32๐‘–๏ป
  • E๏ฌโˆ’12โˆ’12๐‘–,12+12๐‘–๏ธ

Q2:

If ๐‘=๐‘Ÿ(๐œƒ+๐‘–๐œƒ)cossin, what is ๐‘๏Š, where ๐‘› is an integer?

  • A๐‘Ÿ๏€ฝ๐œƒ๐‘›+๐‘–๐œƒ๐‘›๏‰cossin
  • B๐‘Ÿ(๐‘›๐œƒ+๐‘–๐‘›๐œƒ)๏Šcossin
  • C๐‘Ÿ(๐œƒ+๐‘–๐œƒ)๏Šcossin
  • D๐‘Ÿ(๐‘›๐œƒ+๐‘–๐‘›๐œƒ)cossin

Q3:

What is (1+๐‘–)๏Šง๏Šฆ?

  • A1+๐‘–
  • B2+2๐‘–
  • C10๐‘–
  • D32๐‘–
  • E2

Q4:

Use De Moivreโ€™s theorem to find the two square roots of cossin๐œ‹3+๐‘–๐œ‹3.

  • A๏ฏโˆš32+12๐‘–,๐‘–๏ป
  • B๏ฏ12โˆ’โˆš32๐‘–,โˆ’12+โˆš32๐‘–๏ป
  • C{๐‘–,โˆ’๐‘–}
  • D๏ฏโˆš32+12๐‘–,โˆ’โˆš32โˆ’12๐‘–๏ป
  • E๏ฌ12+โˆš2๐‘–,โˆ’12โˆ’โˆš2๐‘–๏ธ

Q5:

Use De Moivreโ€™s theorem to find the two square roots of 9๏€ป๐œ‹3+๐‘–๐œ‹3๏‡cossin.

  • A๏ฏ3โˆš32+32๐‘–,3๐‘–๏ป
  • B{โˆ’1,1}
  • C{3๐‘–,โˆ’3๐‘–}
  • D๏ฏ3โˆš32+32๐‘–,โˆ’3โˆš32โˆ’32๐‘–๏ป
  • E๏ฌ12+โˆš2๐‘–,โˆ’12โˆ’โˆš2๐‘–๏ธ

Q6:

What is (โˆ’1+๐‘–)๏Šฎ?

  • Aโˆ’1+๐‘–
  • Bโˆ’8+8๐‘–
  • Cโˆ’8๐‘–
  • D16
  • E2

Q7:

Given that ๐‘ฅ+๐‘ฆ๐‘–=๏€ผ1โˆ’1๐‘–๏ˆ๏Šฌ, where ๐‘ฅ and ๐‘ฆ are real numbers, determine the value of ๐‘ฅ and the value of ๐‘ฆ.

  • A๐‘ฅ=0, ๐‘ฆ=โˆ’8
  • B๐‘ฅ=0, ๐‘ฆ=8
  • C๐‘ฅ=0, ๐‘ฆ=โˆ’2
  • D๐‘ฅ=0, ๐‘ฆ=2

Q8:

Simplify 18(โˆ’๐‘–+1)(๐‘–+1)๏Šฉ๏Šฏ๏Šช๏Šง.

Q9:

Consider the complex number ๐‘ง=3โˆ’๐‘–.

Find the modulus of ๐‘ง.

  • Aโˆš2
  • Bโˆš8
  • C1
  • D3
  • Eโˆš10

Hence, find the modulus of ๐‘ง๏Šซ.

  • A10
  • B100โˆš10
  • C243
  • Dโˆš10
  • E10โˆš10

Q10:

Determine, in trigonometric form, the square roots of ๏€ฝโˆ’5โˆ’5๐‘–โˆ’5+5๐‘–๏‰๏Šฏ.

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

Q11:

Simplify (โˆ’1โˆ’๐‘–)๏Šฌ, giving your answer in trigonometric form.

  • A8(180+๐‘–180)sincosโˆ˜โˆ˜
  • B8(90+๐‘–90)cossinโˆ˜โˆ˜
  • Cโˆ’8(270+๐‘–270)cossinโˆ˜โˆ˜
  • D8(270+๐‘–270)cossinโˆ˜โˆ˜
  • E8(270+๐‘–270)sincosโˆ˜โˆ˜

Q12:

Consider the complex number ๐‘ง=1+โˆš3๐‘–.

Find the modulus of ๐‘ง.

Find the argument of ๐‘ง.

  • A2
  • B๐œ‹6
  • C๐œ‹3
  • Dโˆš10
  • E2๐œ‹3

Hence, use the properties of multiplication of complex numbers in polar form to find the modulus and argument of ๐‘ง๏Šฉ.

  • Amodulus = 8, argument = ๐œ‹
  • Bmodulus = โˆš10, argument = ๐œ‹
  • Cmodulus = 8, argument = ๐œ‹2
  • Dmodulus = โˆš3, argument= ๐œ‹
  • Emodulus = โˆš10, argument = ๐œ‹2

Hence, find the value of ๐‘ง๏Šฉ.

Q13:

Given that ๐‘ง=7(315+๐‘–315)sincosโˆ˜โˆ˜, find ๐‘ง๏Šจ, giving your answer in exponential form.

  • A7๐‘’๏Žข๏‘ฝ๏Žก๏ƒ
  • B49๐‘’๏Žข๏‘ฝ๏Žฃ๏ƒ
  • C14๐‘’๏Žข๏‘ฝ๏Žก๏ƒ
  • D49๐‘’๏Žฆ๏‘ฝ๏Žฃ๏ƒ
  • E49๐‘’๏Žข๏‘ฝ๏Žก๏ƒ

Q14:

Given that ๐‘ง=3โˆš2(225โˆ’๐‘–225)cossinโˆ˜โˆ˜, find ๐‘ง๏Šจ, giving your answer in exponential form.

  • A18๐‘’๏Žข๏‘ฝ๏Žก๏ƒ
  • B18๐‘’๏Žข๏‘ฝ๏Žฃ๏ƒ
  • C6โˆš2๐‘’๏Žข๏‘ฝ๏Žก๏ƒ
  • D3โˆš2๐‘’๏Žข๏‘ฝ๏Žก๏ƒ

Q15:

If ๐‘ง=6(225+๐‘–225)๏Šงโˆ˜โˆ˜cossin, ๐‘ง=90+๐‘–90๏Šจโˆ˜โˆ˜cossin, and ๐‘ง=270+๐‘–270๏Šฉโˆ˜โˆ˜cossin, what is the exponential form of (๐‘ง๐‘ง๐‘ง)๏Šง๏Šจ๏Šฉ๏Šจ?

  • A36๐‘’๏Žข๏‘ฝ๏Žก๏ƒ
  • B36๐‘’๏‘ฝ๏Žก๏ƒ
  • C36๐‘’๏Žค๏‘ฝ๏Žฃ๏ƒ
  • D6๐‘’๏‘ฝ๏Žก๏ƒ
  • E6๐‘’๏Žก๏‘ฝ๏Žข๏ƒ

Q16:

Given that ๐‘ง=2โˆš3(240+๐‘–240)cossinโˆ˜โˆ˜, find ๐‘ง๏Šจ in exponential form.

  • A๐‘ง=12๐‘’๏Šจ๏ƒ๏Žฃ๏‘ฝ๏Žข
  • B๐‘ง=2โˆš3๐‘’๏Šจ๏ƒ๏Žก๏‘ฝ๏Žข
  • C๐‘ง=12๐‘’๏Šจ๏ƒ๏Žฆ๏‘ฝ๏Žฅ
  • D๐‘ง=4โˆš3๐‘’๏Šจ๏ƒ๏Žก๏‘ฝ๏Žข
  • E๐‘ง=12๐‘’๏Šจ๏ƒ๏Žก๏‘ฝ๏Žข

Q17:

Find the possible values of ๏€ฟโˆ’27โˆš32โˆ’27๐‘–2๏‹๏Žฃ๏Žข, giving your answers in trigonometric form.

  • A81(40+๐‘–40)cossinโˆ˜โˆ˜, 81(160+๐‘–160)cossinโˆ˜โˆ˜, 81(280+๐‘–280)cossinโˆ˜โˆ˜
  • B81(70+๐‘–70)cossinโˆ˜โˆ˜, 81(190+๐‘–190)cossinโˆ˜โˆ˜, 81(310+๐‘–310)cossinโˆ˜โˆ˜
  • C27(40+๐‘–40)cossinโˆ˜โˆ˜, 27(160+๐‘–160)cossinโˆ˜โˆ˜, 27(280+๐‘–280)cossinโˆ˜โˆ˜
  • D27, 27(120+๐‘–120)cossinโˆ˜โˆ˜, 27(240+๐‘–240)cossinโˆ˜โˆ˜
  • E81, 81(120+๐‘–120)cossinโˆ˜โˆ˜, 81(240+๐‘–240)cossinโˆ˜โˆ˜

Q18:

If ๐‘ง=3(45+๐‘–45)cossinโˆ˜โˆ˜, what is ๐‘ง๏Šจ?

  • A9(90+๐‘–90)cossinโˆ˜โˆ˜
  • B6๏€บ45+๐‘–45๏†cossin๏Šจโˆ˜๏Šจโˆ˜
  • C6(90+๐‘–90)cossinโˆ˜โˆ˜
  • D3๏€บ45+๐‘–45๏†cossin๏Šจโˆ˜๏Šจโˆ˜
  • E9(45+๐‘–45)cossinโˆ˜โˆ˜

Q19:

Given that ๐‘ง=โˆš32โˆ’32๐‘–, find ๐‘ง๏Šซ, giving your answer in exponential form.

  • A9โˆš3๐‘’๏‘ฝ๏Žฅ๏ƒ
  • B9โˆš3๐‘’๏‘ฝ๏Žข๏ƒ
  • Cโˆš3๐‘’๏‘ฝ๏Žข๏ƒ
  • D5โˆš3๐‘’๏‘ฝ๏Žข๏ƒ

Q20:

Given that ๐‘ง=8(240+๐‘–240)๏Šงโˆ˜โˆ˜cossin, ๐‘ง=4๏€ผ5๐œ‹4+๐‘–5๐œ‹4๏ˆ๏Šจcossin, and ๐‘ง=8(45+๐‘–45)๏Šฉโˆ˜โˆ˜cossin, find ๐‘ง๐‘ง๐‘ง๏Šง๏Šฌ๏Šจ๏Šช๏Šฉ, giving your answer in exponential form.

  • A8๐‘’๏Ž ๏Ž ๏‘ฝ๏Žฅ๏ƒ
  • B32,768๐‘’๏Ž ๏Ž ๏‘ฝ๏Žฅ๏ƒ
  • C8๐‘’๏‘ฝ๏Žข๏ƒ
  • D8๐‘’๏Žค๏‘ฝ๏Žฅ๏ƒ
  • E4๐‘’๏Ž ๏Ž ๏‘ฝ๏Žฅ๏ƒ

Q21:

Given that ๐‘=8๏€ผ๏€ผ19๐œ‹12๏ˆโˆ’๐‘–๏€ผ19๐œ‹12๏ˆ๏ˆ๏Šง๏Šจcossin and ๐‘=3๐‘’๏Šจ๏ƒ๏Ž ๏Ž ๏‘ฝ๏Žฅ, where ๐‘–=โˆ’1๏Šจ, express ๐‘=๐‘๐‘๏Šง๏Šจ๏Šจ in trigonometric form.

  • A๐‘=๏€ป๏€ป๐œ‹2๏‡+๐‘–๏€ป๐œ‹2๏‡๏‡cossin
  • B๐‘=72๏€ป๏€ป๐œ‹2๏‡+๐‘–๏€ป๐œ‹2๏‡๏‡cossin
  • C๐‘=72๏€ป๏€ปโˆ’๐œ‹2๏‡+๐‘–๏€ปโˆ’๐œ‹2๏‡๏‡cossin
  • D๐‘=๏€ป๏€ปโˆ’๐œ‹2๏‡+๐‘–๏€ปโˆ’๐œ‹2๏‡๏‡cossin

Q22:

Given that ๐‘=๏€ปโˆš3โˆ’๐‘–๏‡๏Š and |๐‘|=32, determine the principal amplitude of ๐‘.

  • A๐œ‹3
  • B๐œ‹2
  • C๐œ‹6
  • Dโˆ’5๐œ‹6

Q23:

Given that ๐‘=โˆ’30+30๐‘–, determine the principal amplitude of ๐‘๏Šซ.

Q24:

Consider the equation ๐‘ง=2โˆš3+2๐‘–๏Šฉ.

Express 2โˆš3+2๐‘– in polar form using the general form of the argument.

  • A2โˆš3+2๐‘–=4๏€ฝ๏€ฝ๐œ‹+2๐œ‹๐‘˜3๏‰+๐‘–๏€ฝ๐œ‹+2๐œ‹๐‘˜3๏‰๏‰cossin for ๐‘˜โˆˆโ„ค
  • B2โˆš3+2๐‘–=4๏€ฝ๏€ฝ๐œ‹+2๐œ‹๐‘˜6๏‰+๐‘–๏€ฝ๐œ‹+2๐œ‹๐‘˜6๏‰๏‰cossin for ๐‘˜โˆˆโ„ค
  • C2โˆš3+2๐‘–=4๏€ป๏€ป๐œ‹6+๐œ‹๐‘˜๏‡+๐‘–๏€ป๐œ‹6+๐œ‹๐‘˜๏‡๏‡cossin for ๐‘˜โˆˆโ„ค
  • D2โˆš3+2๐‘–=4๏€ป๏€ป๐œ‹6+2๐œ‹๐‘˜๏‡+๐‘–๏€ป๐œ‹6+2๐œ‹๐‘˜๏‡๏‡cossin for ๐‘˜โˆˆโ„ค
  • E2โˆš3+2๐‘–=4๏€ป๏€ป๐œ‹3+๐œ‹๐‘˜๏‡+๐‘–๏€ป๐œ‹3+๐œ‹๐‘˜๏‡๏‡cossin for ๐‘˜โˆˆโ„ค

By applying de Moivreโ€™s theorem to the left-hand side, rewrite the equation in polar form.

  • A๐‘Ÿ(3๐œƒ+๐‘–3๐œƒ)=4๏€ป๏€ป๐œ‹6+2๐œ‹๐‘˜๏‡+๐‘–๏€ป๐œ‹6+2๐œ‹๐‘˜๏‡๏‡๏Šฉcossincossin
  • B๐‘Ÿ(3๐œƒ+๐‘–3๐œƒ)=4๏€ฝ๏€ฝ๐œ‹+2๐œ‹๐‘˜6๏‰+๐‘–๏€ฝ๐œ‹+2๐œ‹๐‘˜6๏‰๏‰๏Šฉcossincossin
  • C๐‘Ÿ(๐œƒ+๐‘–๐œƒ)=4๏€ป3๏€ป๐œ‹6+2๐œ‹๐‘˜๏‡+๐‘–3๏€ป๐œ‹6+2๐œ‹๐‘˜๏‡๏‡cossincossin๏Šฉ
  • D๐‘Ÿ(๐œƒ+๐‘–๐œƒ)=โˆš4๏€ฝ๏€ฝ๐œ‹+2๐œ‹๐‘˜18๏‰+๐‘–๏€ฝ๐œ‹+2๐œ‹๐‘˜18๏‰๏‰cossincossin๏Žข
  • E๐‘Ÿ(๐œƒ+๐‘–๐œƒ)=โˆš4๏€ฝ๏€ฝ๐œ‹+2๐œ‹๐‘˜2๏‰+๐‘–๏€ฝ๐œ‹+2๐œ‹๐‘˜2๏‰๏‰cossincossin๏Žข

By equating the moduli and arguments and considering different values of the general argument, find the 3 cube roots of 2โˆš3+2๐‘–, expressing them in exponential form.

  • A๐‘ง=โˆš4๐‘’๏Žข๏‘ฝ๏Ž ๏Žง๏ƒ, ๏Žข๏Ž ๏Žข๏‘ฝ๏Ž ๏Žงโˆš4๐‘’๏ƒ, ๏Žข๏Ž ๏Ž ๏‘ฝ๏Ž ๏Žงโˆš4๐‘’๏Šฑ๏ƒ
  • B๐‘ง=โˆš4๐‘’๏Žข๏Žก๏‘ฝ๏Žจ๏ƒ, ๏Žข๏Žง๏‘ฝ๏Žจโˆš4๐‘’๏ƒ, ๏Žข๏Žฃ๏‘ฝ๏Žจโˆš4๐‘’๏Šฑ๏ƒ
  • C๐‘ง=โˆš4๐‘’๏Žข๏‘ฝ๏Žจ๏ƒ, ๏Žข๏Žฆ๏‘ฝ๏Žจโˆš4๐‘’๏ƒ, ๏Žข๏Žค๏‘ฝ๏Žจโˆš4๐‘’๏Šฑ๏ƒ
  • D๐‘ง=โˆš4๐‘’๏Žข๏‘ฝ๏Ž ๏Žง๏ƒ, ๏Žข๏Žข๏‘ฝ๏Ž ๏Žงโˆš4๐‘’๏ƒ, ๏Žข๏Žฅ๏‘ฝ๏Ž ๏Žงโˆš4๐‘’๏ƒ
  • E๐‘ง=โˆš4๐‘’๏Žข๏‘ฝ๏Žก๏ƒ, ๏Žขโˆš4๐‘’๏ƒ๏Ž„, ๏Žข๏‘ฝ๏Žกโˆš4๐‘’๏Šฑ๏ƒ

Q25:

Given that ๐‘=5โˆš22โˆ’5โˆš22๐‘–, find 1๐‘, giving your answer in trigonometric form.

  • A1๐‘=15๏”๏€ผ3๐œ‹4๏ˆ+๐‘–๏€ผ3๐œ‹4๏ˆ๏ cossin
  • B1๐‘=15๏“๏€ป๐œ‹4๏‡+๐‘–๏€ป๐œ‹4๏‡๏Ÿcossin
  • C1๐‘=125๏”๏€ผ5๐œ‹4๏ˆ+๐‘–๏€ผ5๐œ‹4๏ˆ๏ cossin
  • D1๐‘=5๏“๏€ป๐œ‹4๏‡+๐‘–๏€ป๐œ‹4๏‡๏Ÿcossin
  • E1๐‘=125๏“๏€ป๐œ‹4๏‡+๐‘–๏€ป๐œ‹4๏‡๏Ÿcossin

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