Question

If (-1/2+ i√3/2)^1000 = a+ib, find the values of a and b​

Answer 1

SOLUTION

GIVEN

$\displaystyle \sf{ { \bigg( - \frac{1}{2} + i \frac{ \sqrt{3} }{2} \bigg)}^{1000} = a + ib}$

TO DETERMINE

The value of a and b

CONCEPT TO BE IMPLEMENTED

Complex Number

A complex number z = a + ib is defined as an ordered pair of Real numbers ( a, b) that satisfies the following conditions :

(i) Condition for equality :

(a, b) = (c, d) if and only if a = c, b = d

(ii) Definition of addition :

(a, b) + (c, d) = (a+c, b+ d)

(iii) Definition of multiplication :

(a, b). (c, d) = (ac-bd , ad+bc )

Of the ordered pair (a, b) the first component a is called Real part of z and the second component b is called Imaginary part of z

EVALUATION

Here it is given that

$\displaystyle \sf{ { \bigg( - \frac{1}{2} + i \frac{ \sqrt{3} }{2} \bigg)}^{1000} = a + ib}$

First we simplify

$\displaystyle \sf{ { \bigg( - \frac{1}{2} + i \frac{ \sqrt{3} }{2} \bigg)}^{1000} }$

$\therefore \: \: \displaystyle \sf{ { \bigg( - \frac{1}{2} + i \frac{ \sqrt{3} }{2} \bigg)}^{1000} }$

$= \displaystyle \sf{ { \bigg( \frac{1}{2} - i \frac{ \sqrt{3} }{2} \bigg)}^{1000} }$

$= \displaystyle \sf{ { \bigg( \cos \frac{\pi}{3} - i \sin \frac{\pi}{3} \bigg)}^{1000} }$

$\displaystyle \sf{ = { \bigg( \cos \frac{1000\pi}{3} - i \sin \frac{1000\pi}{3} \bigg)} } \: \: Using De Moivre's Theorem$

$\displaystyle \sf{ = { \bigg( \cos {60000}^{ \circ} - i \sin {60000}^{ \circ} \bigg)} }$

= $\displaystyle \sf{ - \frac{1}{2} + i \frac{ \sqrt{3} }{2} }$

$\therefore \: \: \displaystyle \sf{ - \frac{1}{2} + i \frac{ \sqrt{3} }{2} = a + ib }$

Comparing we get

$\displaystyle \sf{ a = - \frac{1}{2} \: \: and \: \: b = \frac{ \sqrt{3} }{2} }$

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