Verified Solution • 2025–26 Session moderate

QJEEM23Ma7970: JEE Main 2023 Mathematics - Class 11 JEEM23Ma7970 Solution: The complex number is equal to :

The complex number z=i−1cos⁡π3+isin⁡π3z=\frac{i-1}{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}} is equal to :

Multiple Choice Options

  • A cos⁡π12−isin⁡π12\cos \frac{\pi}{12}-i \sin \frac{\pi}{12}
  • B 2(cos⁡π12+isin⁡π12)\sqrt{2}\left(\cos \frac{\pi}{12}+i \sin \frac{\pi}{12}\right)
  • C 2i(cos⁡5π12−isin⁡5π12)\sqrt{2} i\left(\cos \frac{5 \pi}{12}-i \sin \frac{5 \pi}{12}\right)
  • D 2(cos⁡5π12+isin⁡5π12)\sqrt{2}\left(\cos \frac{5 \pi}{12}+i \sin \frac{5 \pi}{12}\right)

Step-by-Step Verified Solution

Z=i−1cos⁡π3+isin⁡π3=i−112+32i\mathrm{Z}=\frac{\mathrm{i}-1}{\cos \frac{\pi}{3}+\mathrm{i} \sin \frac{\pi}{3}}=\frac{\mathrm{i}-1}{\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}}

=i−112+32i×12−32i12−3/2i=3−12+3+12i=\frac{i-1}{\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}} \times \frac{\frac{1}{2}-\sqrt{\frac{3}{2}} \mathrm{i}}{\frac{1}{2}-\sqrt{3 / 2} \mathrm{i}}=\frac{\sqrt{3}-1}{2}+\frac{\sqrt{3}+1}{2} \mathrm{i}

Apply polar form,

rcos⁡θ=3−12r \cos \theta=\frac{\sqrt{3}-1}{2}

rsin⁡θ=3+12r \sin \theta=\frac{\sqrt{3}+1}{2}

Now, tan⁡θ=3+13−1\tan \theta=\frac{\sqrt{3}+1}{\sqrt{3}-1}

So, θ=5π12\theta=\frac{5 \pi}{12}

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