250+ TOP MCQs on DeMoivre’s Theorem and Answers

Complex Analysis Multiple Choice Questions on “DeMoivre’s Theorem”.

1. Find the value of (1+i)¹⁰⁰.
a) 2¹⁰⁰ (cos⁡100π+isin100π)
b) 2¹⁰⁰ (cos⁡25π+isin25π)
c) 2⁵⁰ (cos⁡100π+isin100π)
d) 2⁵⁰ (cos⁡25π+isin25π)
View Answer

Answer: d
Explanation: We know that,
1+i=(sqrt 2 (frac{1}{sqrt 2}+frac{i}{sqrt 2})=sqrt 2 (cos frac{pi}{4}+isin frac{pi}{4}))
((1+i)^{100}=(sqrt 2 left (cos frac{pi}{4}+isin frac{pi}{4}right ))^{100}=2^{50}(left(cos frac{pi}{4}+isin frac{pi}{4}right))^{100})
By Applying the DeMoivre’s Theorem
((1+i)^{100}=2^{50} left (cos 100frac{pi}{4}+isin 100frac{pi}{4} right ))
((1+i)^{100}=2^{50} (cos⁡25pi+isin25pi)).

2. Find the value of (1-i)¹⁰⁰.
a) 2¹⁰⁰ (cos⁡100π-isin100π)
b) 2¹⁰⁰ (cos⁡25π-isin25π)
c) 2⁵⁰ (cos⁡25π-isin25π)
d) 2⁵⁰ (cos⁡100π-isin100π)
View Answer

Answer: c
Explanation: We know that,
(1-i=sqrt 2 (frac{1}{sqrt 2}-frac{i}{sqrt 2})=sqrt 2 (cos frac{pi}{4}-isin frac{pi}{4}))
((1-i) ^{100}=(sqrt 2 (cos frac{pi}{4}-isin frac{pi}{4})) ^{100}=2^{50}((cos frac{pi}{4}-isin frac{pi}{4})) ^{100})
By Applying the DeMoivre’s Theorem
((1-i)^{100}=2^{50} (cos frac{100pi}{4}-isin frac{100pi}{4}))
((1-i)^{100}=2^{50} (cos⁡25pi-sin25pi)).

3. If (a=frac{1}{sqrt 2}+frac{i}{sqrt 2}), find the value of a⁵ + conjugate of a⁵=?
a) (cos⁡ frac{3pi}{4})
b) (2 sin ⁡frac{5pi}{4})
c) (2 cos frac{5pi}{4})
d) (cos frac{5pi}{4})
View Answer

Answer: c
Explanation: We know that,
(a=(frac{1}{sqrt 2}+frac{i}{sqrt 2})=(cos frac{pi}{4}+isin frac{pi}{4}))
Conjugate of a = ((frac{1}{sqrt 2}-frac{i}{sqrt 2})=(cos frac{pi}{4}-isin frac{pi}{4}))
(a^5) + (conj of a)⁵=((cos frac{pi}{4}+isin frac{pi}{4})^5+(cos frac{pi}{4}-isin frac{pi}{4})^5)
By Applying the DeMoivre’s Theorem
a⁵ + (conj of a)⁵ = (cos⁡ frac{5pi}{4} +isin frac{5pi}{4}+cos⁡ frac{5pi}{4} -isin frac{5pi}{4})
a⁵ + (conj of a)⁵ = (2 cos ⁡frac{5pi}{4}).

4. Evaluate (frac{(1+sqrt 3 i) ^{16}}{(sqrt 3-i) ^{17}}).
a) (frac{1}{2}×(cos frac{pi}{6}+isin frac{pi}{6}))
b) (frac{1}{3}×(cos frac{pi}{6}+isin frac{pi}{6}))
c) (frac{1}{2}×(cos frac{pi}{3}+isin frac{pi}{3}))
d) (frac{1}{4}×(cos frac{pi}{6}+isin frac{pi}{6}))
View Answer

Answer: a
Explanation: We know that,
(1+sqrt 3 i=2(cos frac{pi}{3}+isin frac{pi}{3}))
(sqrt 3-i=2(cos frac{pi}{6}-isin frac{pi}{6}))
(frac{(1+sqrt 3 i) ^{16}}{(sqrt 3-i) {^17}}=frac{2^{16}(cos frac{pi}{3}+isin frac{pi}{3}) ^{16}}{2^{17}(cos frac{pi}{6}-isin frac{pi}{6}) ^{17}})
By Demoivre’s Theorem
= (frac{1}{2}×(cos frac{16pi}{3}+isin frac{16pi}{3})(cos frac{17pi}{6}+isin frac{17pi}{6}))
= (frac{1}{2}×(cos(frac{⁡49pi}{6})+isin(frac{49pi}{6})))
= (frac{1}{2}×(cosfrac{pi}{6}+isinfrac{pi}{6})).

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