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,
\[\begin{align}
  & \Rightarrow 4+8\left( -\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right) \\
 & \Rightarrow 4-4+4\sqrt{3}i=4\sqrt{3}i \\
\end{align}\]
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 \[\sin (-\theta )=-\sin \theta \].