Complex Numbers Question and Answers – DeMoivre’s Theorem

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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 (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
\((1+i)^{100}=2^{50} \left (cos 100\frac{\pi}{4}+isin 100\frac{\pi}{4} \right )\)
\((1+i)^{100}=2^{50} (cos⁡25\pi+isin25\pi)\).
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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
\((1-i)^{100}=2^{50} (cos \frac{100\pi}{4}-isin \frac{100\pi}{4})\)
\((1-i)^{100}=2^{50} (cos⁡25\pi-sin25\pi)\).

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

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 = \(cos⁡ \frac{5\pi}{4} +isin \frac{5\pi}{4}+cos⁡ \frac{5\pi}{4} -isin \frac{5\pi}{4}\)
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

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
= \(\frac{1}{2}×(cos \frac{16\pi}{3}+isin \frac{16\pi}{3})(cos \frac{17\pi}{6}+isin \frac{17\pi}{6})\)
= \(\frac{1}{2}×(cos(\frac{⁡49\pi}{6})+isin(\frac{49\pi}{6}))\)
= \(\frac{1}{2}×(cos\frac{\pi}{6}+isin\frac{\pi}{6})\).
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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn