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Exercise 1.3 · Q99

Q.Express the following complex number in polar form and exponential form : 1+7i(2−i)2\dfrac{1+7i}{(2-i)^2}

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(2−i)2=4−4i+i2=3−4i(2-i)^2=4-4i+i^2=3-4i. So the expression is dfrac1+7i3−4i\\dfrac{1+7i}{3-4i}. Rationalise: dfrac(1+7i)(3+4i)(3−4i)(3+4i)=dfrac3+4i+21i+28i29+16=dfrac3+25i−2825=dfrac−25+25i25=−1+i\\dfrac{(1+7i)(3+4i)}{(3-4i)(3+4i)}=\\dfrac{3+4i+21i+28i^2}{9+16}=\\dfrac{3+25i-28}{25}=\\dfrac{-25+25i}{25}=-1+i. So a=−1,b=1a=-1,b=1: ∣z∣=sqrt2|z|=\\sqrt2. Quadrant II: argz=tan−1(−1)+pi=dfrac3pi4\\arg z=\\tan^{-1}(-1)+\\pi=\\dfrac{3\\pi}{4}. So $z=\sqrt2\left(\cos\dfrac{3\pi}{4}+i\sin\dfrac{3\pi}{4}\right)= …

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