## What are we learning ?

**Competency in Focus:** Complex Number

This problem from B.Stat 2005 Problem 4 – Objective Admission Test (Hons.) is based on Demoivre’s Theorem.## First look at the knowledge graph:-

## Next understand the problem

The value of $\left\{\frac{1}{2}(-1+\sqrt{3} i)\right\}^{15}+\left\{\frac{1}{2}(-1-\sqrt{3} i)\right\}^{15}$ is

$\begin{array}{ll}{\text { (A) }-1 } & {\text { (B) } 0 \text { (C) } \frac{1}{2^{14}} \quad(\mathrm {D} ) 2}\end{array}$

$\begin{array}{ll}{\text { (A) }-1 } & {\text { (B) } 0 \text { (C) } \frac{1}{2^{14}} \quad(\mathrm {D} ) 2}\end{array}$

##### Source of the problem

B.Stat 2005 Problem 4 – Objective Admission Test (Hons.)

##### Key Competency

### Complex Number

##### Difficulty Level

4/10

##### Suggested Book

Challenges and Thrills in Pre College Mathematics Excursion Of Mathematics

## Start with hints

Do you really need a hint? Try it first!

First of all know the Theorem that we are going to use. its called De-Moivres Theorem. See below $(\cos \theta+i \sin \theta)^{n}=\cos n \theta+i \sin n \theta$

for all integers $n$.

for all integers $n$.

I hope you must have solved it, becouse it just involves basic calculation and trignometric solutions. If not then see Next hint.

$\left\{\frac{1}{2}(-1+\sqrt{3} i)\right\}^{15}+\left\{\frac{1}{2}(-1-\sqrt{3} i)\right\}^{15}$

$=\left(\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}\right) \quad+\left(\cos \frac{4 \pi}{3}+i \sin \frac{4 \pi}{3}\right)$

$=\cos \frac{2 \times 15 \times \pi}{3}+i \sin \frac{2 \times 15 \times \pi}{3}+\cos \frac{4 \times 15 \times \pi}{3}+i \sin \frac{4 \times 15 \times \pi}{3}$

$=\cos 10 \pi+i \sin 10 \pi+\cos 20 \pi+i \sin 20 \pi$

$=2$

$=\left(\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}\right) \quad+\left(\cos \frac{4 \pi}{3}+i \sin \frac{4 \pi}{3}\right)$

$=\cos \frac{2 \times 15 \times \pi}{3}+i \sin \frac{2 \times 15 \times \pi}{3}+\cos \frac{4 \times 15 \times \pi}{3}+i \sin \frac{4 \times 15 \times \pi}{3}$

$=\cos 10 \pi+i \sin 10 \pi+\cos 20 \pi+i \sin 20 \pi$

$=2$

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