AHL 1.13—Complex numbers continued

For $z=-\sqrt{3}+i$, we have $a=-\sqrt{3}$ and $b=1$. The modulus is $r=\sqrt{(-\sqrt{3}{)}^2+1^2}=\sqrt{3+1}=\sqrt{4}=2$.

The reference angle is $\arctan \left(\frac{|b|}{|a|}\right)=\arctan \left(\frac{1}{\sqrt{3}}\right)=\frac{\pi }{6}$.

Since $a<0$ and $b>0$, the number lies in the second quadrant. The principal argument is $\theta =\pi -\frac{\pi }{6}=\frac{5\pi }{6}$.

Therefore $z=2e^{i\cdot 5\pi /6}$, confirming the first option is correct.

We need the correct modulus $r$ and principal argument $\theta$ for $z=-\sqrt{3}+i$. The modulus is clearly $r=2$, so any option with $\sqrt{2}$ as the modulus can be eliminated immediately.

The modulus of $z$ is $\sqrt{3+1}=2$, not $\sqrt{2}$. So $\sqrt{2}{e}^{i\cdot 5\pi /6}$ is wrong.

The argument $\pi /6$ corresponds to a first-quadrant number. However $-\sqrt{3}+i$ has a negative real part, placing it in the second quadrant, so $\pi /6$ is incorrect.

The argument $2\pi /3$ gives $\cos (2\pi /3)=-1/2$ and $\sin (2\pi /3)=\sqrt{3}/2$, yielding $z=2(-1/2+i\sqrt{3}/2)=-1+i\sqrt{3}$, which does not match $-\sqrt{3}+i$.

$2e^{i\cdot 5\pi /6}$ gives $2(\cos (5\pi /6)+i\sin (5\pi /6))=2(-\sqrt{3}/2+i/2)=-\sqrt{3}+i$. ✓

$r=\sqrt{(-2{)}^2+(-2{)}^2}=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}$.

$\arctan \left(\frac{|-2|}{|-2|}\right)=\arctan (1)=\frac{\pi }{4}$.

Since $a<0$ and $b<0$, $z$ lies in the third quadrant. The principal argument is $\theta =\frac{\pi }{4}-\pi =-\frac{3\pi }{4}$.

$r=2\sqrt{2}$ and $\theta =-\frac{3\pi }{4}$, giving $z=2\sqrt{2}\mathrm{cis}\left(-\frac{3\pi }{4}\right)$.

We need $r$ and $\theta$ for $z=-2-2i$. The modulus is $\sqrt{4+4}=2\sqrt{2}$, so any answer with $r=4$ is wrong.

The option $r=4,\theta =-3\pi /4$ has the wrong modulus since $\sqrt{(-2{)}^2+(-2{)}^2}=2\sqrt{2}\ne 4$.

$\theta =3\pi /4$ places $z$ in the second quadrant (positive imaginary part), but $-2-2i$ has a negative imaginary part. So $r=2\sqrt{2},\theta =3\pi /4$ is wrong.

$\theta =\pi /4$ places $z$ in the first quadrant, giving $z$ with positive real and imaginary parts. This contradicts $-2-2i$.

The only remaining option is $r=2\sqrt{2},\theta =-\frac{3\pi }{4}$, which correctly represents the third-quadrant point $-2-2i$.