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Miscellaneous Exercise 1 (Subjective) · Q196

Q.Convert the complex number in polar form and also in exponential form : z=2+63 i5+3 iz = \dfrac{2+6\sqrt3\,i}{5+\sqrt3\,i}

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z=dfrac2+6sqrt3i5+sqrt3i=dfrac(2+6sqrt3i)(5−sqrt3i)(5+sqrt3i)(5−sqrt3i)=dfrac10−2sqrt3i+30sqrt3i−6sqrt3cdotsqrt3i225+3=dfrac10+28sqrt3i+1828=dfrac28+28sqrt3i28=1+sqrt3iz=\\dfrac{2+6\\sqrt3i}{5+\\sqrt3i}=\\dfrac{(2+6\\sqrt3i)(5-\\sqrt3i)}{(5+\\sqrt3i)(5-\\sqrt3i)}=\\dfrac{10-2\\sqrt3i+30\\sqrt3i-6\\sqrt3\\cdot\\sqrt3i^2}{25+3}=\\dfrac{10+28\\sqrt3i+18}{28}=\\dfrac{28+28\\sqrt3i}{28}=1+\\sqrt3i. Now ∣1+sqrt3i∣=sqrt1+3=2|1+\\sqrt3i|=\\sqrt{1+3}=2, Quadrant I: arg=tan−1sqrt3=dfracpi3\\arg=\\tan^{-1}\\sqrt3=\\dfrac{\\pi}{3}. So $z=2\left(\cos\dfrac{\pi}{3}+i\ …

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