# Complex Numbers Question and Answers – DeMoivre’s Theorem

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)

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

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)

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) $$cos \frac{5\pi}{4}$$

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) $$\frac{1}{4}×(cos \frac{\pi}{6}+isin \frac{\pi}{6})$$

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
= $$\frac{1}{2}×(cos \frac{16\pi}{3}+isin \frac{16\pi}{3})(cos \frac{17\pi}{6}+isin \frac{17\pi}{6})$$
= $$\frac{1}{2}×(cos(\frac{⁡49\pi}{6})+isin(\frac{49\pi}{6}))$$
= $$\frac{1}{2}×(cos\frac{\pi}{6}+isin\frac{\pi}{6})$$.