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4.1 complex nubers — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHS4.1 complex nubers1 Mark
MCQ
${\left( {\frac{{1 + \cos \phi + i\sin \phi }}{{1 + \cos \phi - i\sin \phi }}} \right)^n} = $
  • A
    $\cos n\phi - i\sin n\phi $
  • ✓
    $\cos n\phi + i\sin n\phi $
  • C
    $\sin n\phi + i\cos n\phi $
  • D
    $\sin n\phi - i\cos n\phi $
✓

Answer

Correct option: B.
$\cos n\phi + i\sin n\phi $
b
(b)$L.H.S.$ $ = {\left[ {\frac{{2{{\cos }^2}(\phi /2) + 2i\sin (\phi /2)\cos (\phi /2)}}{{2{{\cos }^2}\,(\phi /2) - 2i\sin (\phi /2)\cos (\phi /2)}}} \right]^n}$

$ = {\left[ {\frac{{\cos \,(\phi /2) + i\sin (\phi /2)}}{{\cos (\phi /2) - i\sin (\phi /2)}}} \right]^n}$

$ = {\left[ {\frac{{{e^{i(\phi /2)}}}}{{{e^{ - i(\phi /2)}}}}} \right]^n} = {({e^{i\phi }})^n}$

$ = \cos n\phi  + i\sin n\phi $.

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