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Exercise 1.3 · Q98

Q.Express the following complex number in polar form and exponential form : 1+2i1−3i\dfrac{1+2i}{1-3i}

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dfrac1+2i1−3i=dfrac(1+2i)(1+3i)(1−3i)(1+3i)=dfrac1+3i+2i+6i21+9=dfrac1+5i−610=dfrac−5+5i10=−dfrac12+dfrac12i\\dfrac{1+2i}{1-3i}=\\dfrac{(1+2i)(1+3i)}{(1-3i)(1+3i)}=\\dfrac{1+3i+2i+6i^2}{1+9}=\\dfrac{1+5i-6}{10}=\\dfrac{-5+5i}{10}=-\\dfrac12+\\dfrac12i. So a=−dfrac12,b=dfrac12a=-\\dfrac12,b=\\dfrac12: ∣z∣=dfracsqrt22|z|=\\dfrac{\\sqrt2}{2}. Quadrant II: argz=tan−1(−1)+pi=−dfracpi4+pi=dfrac3pi4\\arg z=\\tan^{-1}(-1)+\\pi=-\\dfrac{\\pi}{4}+\\pi=\\dfrac{3\\pi}{4}. So $z=\dfrac{\sqrt2}{2}\left(\cos\dfrac{3\pi}{4}+i\sin\dfrac{3\pi}{4}\right)=\dfrac{\sqrt2}{ …

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