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Exercise 1.3 · Q107

Q.Convert the complex number z=i−1cos⁡π3+isin⁡π3z = \dfrac{i-1}{\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}} in the polar form.

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Numerator i−1=−1+ii-1=-1+i has modulus sqrt2\\sqrt2 and (Quadrant II) argument dfrac3pi4\\dfrac{3\\pi}{4}, so i−1=sqrt2left(cosdfrac3pi4+isindfrac3pi4right)i-1=\\sqrt2\\left(\\cos\\dfrac{3\\pi}{4}+i\\sin\\dfrac{3\\pi}{4}\\right). The denominator cosdfracpi3+isindfracpi3\\cos\\dfrac{\\pi}{3}+i\\sin\\dfrac{\\pi}{3} already has modulus 11 and argument dfracpi3\\dfrac{\\pi}{3}. Dividing two polar-form numbers divides their moduli and subtracts their arguments: $z=\dfrac{\sqrt2}{1}\left[\cos\left(\dfrac{3\pi}{4}-\dfrac{\pi}{3}\right)+i\sin\left(\dfrac{3\pi}{4}-\dfrac{\pi}{3}\right)\right]=\sqrt2\left(\co …

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