Question

what is the value of (i+√3)^2019+(i-√3)^2019
1)1
2)-1
3)2i
4)-2i​

Answer 1

Answer:

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Answer 2

Given:

(i+√3)^2019+(i-√3)^2019

To Find:

The sum of the above expression.

Solution:

We know , for a complex number  

• z= x + yi , there exist Euler's formula:
• z = r $e^{i\theta}$ = r cos $\theta$ + i r sin $\theta$ ,
• Where r cos $\theta$ = x and r sin $\theta$ = y

• i + √3 = 2 ( i/2 + √3/2) = 2 $e^{i\pi/6 }$,
• as sin π/6 = 1/2 and cos π/6 = √3/2

Similarly,

• i - √3 = 2 ( i/2 - √3/2) = -2 ( √3/2 - i/2 )

• i - √3 =  -2$e^{-i\pi/6}$

Therefore the expression is :

• (i+√3)^2019+(i-√3)^2019 =  $(2e^{i\pi/6})^{ 2019} + (-2e^{-i\pi/6})^{2019}$

• (i+√3)^2019+(i-√3)^2019  = $2^{2019} e^{i\pi2019/6}$ + $(-2)^{2019} e^{-i\pi2019/6}$

We know 2019 = 336x6 + 3

Therefore

• 2019π/6 = 336x6π/6 + 3π/6 = 336π + π/2
• 2019π/6 = 168 x 2π + π/2 = π/2

Therefore,

• (i+√3)^2019+(i-√3)^2019  = $2^{2019}e^{i\pi/2}$ + $(-2)^{2019}e^{-i\pi/2}$

We know, $e^{i\pi/2}$ = i and $e^{-i\pi/2}$ = -i

• (i+√3)^2019+(i-√3)^2019 = $2^{2019}$i - $2^{2019}$(-i) = 2x $2^{2019}$ i = $2^{2020}$i

The value of (i+√3)^2019+(i-√3)^2019 is  $2^{2020}$i