Question

The complex number $z=\frac{i-1}{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}}$ is equal to :

• A. $\sqrt{2} i\left(\cos \frac{5 \pi}{12}-i \sin \frac{5 \pi}{12}\right)$
• B. $\cos \frac{\pi}{12}-i \sin \frac{\pi}{12}$
• C. $\sqrt{2}\left(\cos \frac{\pi}{12}+i \sin \frac{\pi}{12}\right)$
• D. $\sqrt{2}\left(\cos \frac{5 \pi}{12}+i \sin \frac{5 \pi}{12}\right)$

Hint:

To convert a complex number into polar form, use the formula \( r = \sqrt{x^2 + y^2} \) for the modulus and \( \theta = \tan^{-1} \left( \frac{y}{x} \right) \) for the argument, where \( x \) and \( y \) are the real and imaginary parts of the complex number.

Answer 1

The Correct Option is D

$Z=\frac{i-1}{\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}}=\frac{i-1}{\frac{1}{2}+\frac{\sqrt{3}}{2}i}$
$=\frac{i-1}{\frac{1}{2}+\frac{\sqrt{3}}{2}i}\times \frac{\frac{1}{2}-\sqrt{\frac{3}{2}i}}{\frac{1}{2}-\sqrt{3/2}i}=\frac{\sqrt{3}-1}{2}+\frac{\sqrt{3}+1}{2}i$
Apply polar form,
$r\cos \theta =\frac{\sqrt{3}-1}{2}$
$r\sin \theta =\frac{\sqrt{3}+1}{2}$
Now, $\tan \theta =\frac{\sqrt{3}+1}{\sqrt{3}-1}$
So, $\theta =\frac{5\pi }{12}$
So, the correct option is (D) : $\sqrt{2}\left(\cos \frac{5 \pi}{12}+i \sin \frac{5 \pi}{12}\right)$

Answer 2

We are given the complex number: \[ z = \frac{i - 1}{\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}}. \] Step 1: We can simplify this expression by first multiplying both the numerator and denominator by the conjugate of the denominator: \[ z = \frac{(i - 1)(\cos \frac{\pi}{3} - i \sin \frac{\pi}{3})}{(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3})(\cos \frac{\pi}{3} - i \sin \frac{\pi}{3})}. \] The denominator simplifies as follows: \[ (\cos \frac{\pi}{3})^2 + (\sin \frac{\pi}{3})^2 = 1. \] Thus: \[ z = (i - 1)(\cos \frac{\pi}{3} - i \sin \frac{\pi}{3}). \] Step 2: Expanding the numerator: \[ z = i \cos \frac{\pi}{3} - i^2 \sin \frac{\pi}{3} - \cos \frac{\pi}{3} + i \sin \frac{\pi}{3}. \] Since \( i^2 = -1 \), we get: \[ z = \cos \frac{\pi}{3} + i \left( \sin \frac{\pi}{3} - \cos \frac{\pi}{3} \right). \] Step 3: Now, to convert this into polar form, we compute the modulus and argument of the complex number: \[ r = \sqrt{\left( \cos \frac{\pi}{3} \right)^2 + \left( \sin \frac{\pi}{3} - \cos \frac{\pi}{3} \right)^2}. \] This simplifies to: \[ r = \sqrt{2}. \] For the argument \( \theta \), we use: \[ \tan \theta = \frac{\sin \frac{\pi}{3} - \cos \frac{\pi}{3}}{\cos \frac{\pi}{3}} = \frac{\sqrt{3}/2 - 1/2}{1/2} = \frac{\sqrt{3} - 1}{1}. \] Thus, the argument is: \[ \theta = \frac{5\pi}{12}. \] Step 4: Hence, the polar form of the complex number is: \[ z = \sqrt{2} \left( \cos \frac{5\pi}{12} + i \sin \frac{5\pi}{12} \right). \]