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4.1 complex nubers — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHS4.1 complex nubers1 Mark
MCQ
The complex number $z=\frac{i-1}{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}}$ is equal to $.....$
  • ✓
    $\sqrt{2}\left(\cos \frac{5 \pi}{12}+ i \sin \frac{5 \pi}{12}\right)$
  • 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
    $\sqrt{2} i \left(\cos \frac{5 \pi}{12}- i \sin \frac{5 \pi}{12}\right)$
✓

Answer

Correct option: A.
$\sqrt{2}\left(\cos \frac{5 \pi}{12}+ i \sin \frac{5 \pi}{12}\right)$
a
$Z =\frac{ i -1}{\cos \frac{\pi}{3}+ i \sin \frac{\pi}{3}}=\frac{ i -1}{\frac{1}{2}+\frac{\sqrt{3}}{2} i }$

$=\frac{ i -1}{\frac{1}{2}+\frac{\sqrt{3}}{2} i } \times \frac{\frac{1}{2}-\sqrt{\frac{3}{2} i }}{\frac{1}{2}-\sqrt{3 / 2} i }=\frac{\sqrt{3}-1}{2}+\frac{\sqrt{3}+1}{2} i$

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, $\quad \theta=\frac{5 \pi}{12}$

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