Question:

If $ i = \sqrt -1$ then $ 4 + 5 \bigg(-\frac{1}{2}+\frac{i\sqrt 3}{2}\bigg)^{334}+3\bigg(-\frac{1}{2}+\frac{i\sqrt 3}{2}\bigg)^{365}$ is equal to

Updated On: Aug 21, 2023
  • 1-i $\sqrt 3$
  • -1 + i $\sqrt 3$
  • i$\sqrt 3$
  • -i$\sqrt 3$
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The Correct Option is C

Solution and Explanation

If in a complex number a + ib, the ratio a : b is 1: \(\sqrt 3\) 
then it always convert the complex number in \(\omega\) 
.Since, \(\omega =-\frac{1}{2}+\frac{\sqrt 3}{2}i\)
\(\therefore \, 4+5 \bigg(-\frac{1}{2}+\frac{i\sqrt 3}{2}\bigg)^{334}+3\bigg(-\frac{1}{2}+\frac{i\sqrt 3}{2}\bigg)^{365}\)
\(=4+5\omega^{334}+3\omega^{365}\)
\(=4+5.(\omega^3)^{111}.\omega +3.(\omega^3)^{121}.\omega^2\)
\(= 4+5\omega+3\omega^2\ , \ [\because\ \omega^3=1]\)
\(=1+3+2\omega+3(1+\omega+\omega^2)=1+2\omega +3 \times 0\)
\([\because 1+\omega+\omega^2=0]\)
\(=1+(-1+\sqrt 3 i)=\sqrt 3 i\)

So, the correct answer is (C): \(i\sqrt 3\)

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Concepts Used:

Complex Numbers and Quadratic Equations

Complex Number: Any number that is formed as a+ib is called a complex number. For example: 9+3i,7+8i are complex numbers. Here i = -1. With this we can say that i² = 1. So, for every equation which does not have a real solution we can use i = -1.

Quadratic equation: A polynomial that has two roots or is of the degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b and c are the real numbers.