This set of Complex Analysis Multiple Choice Questions & Answers (MCQs) focuses on “DeMoivre’s Theorem”.
1. Find the value of (1+i)100.
a) 2100 (cos100π+isin100π)
b) 2100 (cos25π+isin25π)
c) 250 (cos100π+isin100π)
d) 250 (cos25π+isin25π)
View Answer
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
\((1+i)^{100}=2^{50} \left (cos 100\frac{\pi}{4}+isin 100\frac{\pi}{4} \right )\)
\((1+i)^{100}=2^{50} (cos25\pi+isin25\pi)\).
2. Find the value of (1-i)100.
a) 2100 (cos100π-isin100π)
b) 2100 (cos25π-isin25π)
c) 250 (cos25π-isin25π)
d) 250 (cos100π-isin100π)
View Answer
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
\((1-i)^{100}=2^{50} (cos \frac{100\pi}{4}-isin \frac{100\pi}{4})\)
\((1-i)^{100}=2^{50} (cos25\pi-sin25\pi)\).
3. If \(a=\frac{1}{\sqrt 2}+\frac{i}{\sqrt 2}\), find the value of a5 + conjugate of a5=?
a) \(cos \frac{3\pi}{4}\)
b) \(2 sin \frac{5\pi}{4}\)
c) \(2 cos \frac{5\pi}{4}\)
d) \(cos \frac{5\pi}{4}\)
View Answer
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 = \(cos \frac{5\pi}{4} +isin \frac{5\pi}{4}+cos \frac{5\pi}{4} -isin \frac{5\pi}{4}\)
a5 + (conj of a)5 = \(2 cos \frac{5\pi}{4}\).
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
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
= \(\frac{1}{2}×(cos \frac{16\pi}{3}+isin \frac{16\pi}{3})(cos \frac{17\pi}{6}+isin \frac{17\pi}{6})\)
= \(\frac{1}{2}×(cos(\frac{49\pi}{6})+isin(\frac{49\pi}{6}))\)
= \(\frac{1}{2}×(cos\frac{\pi}{6}+isin\frac{\pi}{6})\).
Sanfoundry Global Education & Learning Series – Complex Analysis.
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