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Exercise: Modulus and Argand Plane · Q14

Q.Find the modulus and the argument of z=−3−iz=-\sqrt3-i.

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✓ Free question

For z=−3−iz=-\sqrt3-i: ∣z∣=(3)2+12=3+1=2|z|=\sqrt{(\sqrt3)^2+1^2}=\sqrt{3+1}=2. Since a=−3<0a=-\sqrt3<0 and b=−1<0b=-1<0, zz lies in Quadrant III. Reference angle: tan⁡−1(13)=π6\tan^{-1}\left(\dfrac{1}{\sqrt3}\right)=\dfrac{\pi}{6}. Using the principal range (−π,π](-\pi,\pi], the Quadrant-III argument is θ=−(π−π6)=−5π6\theta=-\left(\pi-\dfrac{\pi}{6}\right)=-\dfrac{5\pi}{6}. Check: 2(cos⁡(−5π6)+isin⁡(−5π6))=2(−32−i12)=−3−i2\left(\cos\left(-\dfrac{5\pi}6\right)+i\sin\left(-\dfrac{5\pi}6\right)\right)=2\left(-\dfrac{\sqrt3}2-i\dfrac12\right)=-\sqrt3-i, matching zz.

✓Final answer

∣z∣=2|z|=2, arg⁡(z)=−5π6\arg(z)=-\dfrac{5\pi}{6}.

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