Question

The value of $4+5{{\left( -\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right)}^{334}}+3{{\left( -\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2} \right)}^{335}}$ is
(a) $1-i\sqrt{3}$
(b) $-1+i\sqrt{3}$
(c) $4\sqrt{3}i$
(d) $-i\sqrt{3}$

Answer 1

Hint: We will first convert both the complex terms in polar forms using $x+iy=r(\cos \theta +i\sin \theta )$ and we will use the formula ${{(\cos \theta +i\sin \theta )}^{n}}=\cos (n\theta )+i\sin (n\theta )$ to simplify the expression. And then we will use the basic trigonometric identities a few times to get our answer.

Complete step-by-step answer:
As we know, every complex number can be converted into polar form.
$x+iy=r(\cos \theta +i\sin \theta )........(1)$
And also we know that ${{(\cos \theta +i\sin \theta )}^{n}}=\cos (n\theta )+i\sin (n\theta )......(2)$
The expression mentioned in the question is $4+5{{\left( -\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right)}^{334}}+3{{\left( -\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2} \right)}^{335}}.......(3)$
Now first we will convert the terms in sin and cos using equation (1) in equation (3) and hence we get,
$\Rightarrow 4+5{{\left( \cos \left( \pi -\dfrac{\pi }{3} \right)+i\sin \left( \pi -\dfrac{\pi }{3} \right) \right)}^{334}}+3{{\left( \cos \left( -\left( \pi -\dfrac{\pi }{3} \right) \right)-i\sin \left( -\left( \pi -\dfrac{\pi }{3} \right) \right) \right)}^{335}}.......(4)$
Now subtracting the terms in the brackets and simplifying in equation (4) we get,
$\Rightarrow 4+5{{\left( \cos \left( \dfrac{2\pi }{3} \right)+i\sin \left( \dfrac{2\pi }{3} \right) \right)}^{334}}+3{{\left( \cos \left( -\dfrac{2\pi }{3} \right)-i\sin \left( -\dfrac{2\pi }{3} \right) \right)}^{335}}.......(5)$
We will now apply the formula from equation (2) in equation (5) and hence we get,
$\Rightarrow 4+5\left( \cos \left( \dfrac{2\pi }{3}\times 334 \right)+i\sin \left( \dfrac{2\pi }{3}\times 334 \right) \right)+3\left( \cos \left( -\dfrac{2\pi }{3}\times 335 \right)-i\sin \left( -\dfrac{2\pi }{3}\times 335 \right) \right).......(6)$
Now again simplifying all the terms and rearranging in equation (6) we get,
$\Rightarrow 4+5\left( \cos \left( \dfrac{668\pi }{3} \right)+i\sin \left( \dfrac{668\pi }{3} \right) \right)+3\left( \cos \left( -\dfrac{670\pi }{3} \right)-i\sin \left( -\dfrac{670\pi }{3} \right) \right).......(7)$
Now again converting it in terms of $\pi$ in equation (7) and also using the fact that $\cos (-\theta )=\cos \theta$ and also $\sin (-\theta )=-\sin \theta$ and also rearranging, we get,
$\Rightarrow 4+5\left( \cos \left( 222\pi +\dfrac{2\pi }{3} \right)+i\sin \left( 222\pi +\dfrac{2\pi }{3} \right) \right)+3\left( \cos \left( 223\pi +\dfrac{\pi }{3} \right)+i\sin \left( 223\pi +\dfrac{\pi }{3} \right) \right).......(8)$
Simplifying all the terms in equation (8) we get,
$\Rightarrow 4+5\left( \cos \left( \dfrac{2\pi }{3} \right)+i\sin \left( \dfrac{2\pi }{3} \right) \right)+3\left( -\cos \left( \dfrac{\pi }{3} \right)+i\sin \left( \dfrac{\pi }{3} \right) \right).......(9)$
Now substituting the values of basic standard angles of sin and cos in equation (9) we get,
$\Rightarrow 4+5\left( -\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right)+3\left( -\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right).......(10)$
Now adding and subtracting the terms in equation (10) we get,

& \Rightarrow 4+8\left( -\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right) \\\ & \Rightarrow 4-4+4\sqrt{3}i=4\sqrt{3}i \\\ \end{aligned}$$ Hence the correct answer is option (c). Note: Remembering the conversion of complex form into polar form and conversion of terms with powers into simple terms using equation (1) is the key here. Also we in a hurry can make a mistake in solving equation (8) if we fail to change cos and sin terms using $$\cos (-\theta )=\cos \theta $$ and also $$\sin (-\theta )=-\sin \theta $$.