Yes it is in the 4th quadrant.
It is just in the position (sqrt2, -1) where sqrt2 is the real co-ordinate (horiontal), and -1 is the complex co-ordinate (vertical).
\(z=\sqrt2-i\\ z=\sqrt2-1i\\ \text{The modulus is just }\\ |z|=\sqrt{(\sqrt2)^2+(-1)^2}\\ |z|=\sqrt{2+1}\\ |z|=\sqrt{3}\\ \)
The simplest way to do this is to say if
\(z=a+bi \quad then\\ \text{The first quadrant (equivalent) angle will be } \quad atan(|\frac{b}{a}|)\\ z=\sqrt{2}-1i\\ acute\;angle=atan\frac{1}{\sqrt2}\approx 35.26^\circ\\ \text{The 4th quadrant correct answer will be}\quad \\ \theta\approx 360- 35.26^\circ \approx 324.74^\circ \)
OR you could go the long way as done below.
\(now\\ z=\sqrt{3} \left( \sqrt{\frac{2}{3}}+\frac{-1}{\sqrt{3}} i \; \right)\\ z=r(cos\theta +isin\theta)\\ cos\theta=\sqrt\frac{2}{3}\qquad sin\theta = \frac{-1}{\sqrt{3}}\\ 4th \;\;quad\\ \theta=360-acos(\sqrt\frac{2}{3})\\ \theta \approx 360-35.26^\circ\\ \theta \approx 324.74^\circ\\ \)
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