Hi good people!,

How would I go about finding the modulus, the argument and plotting: \(z=\sqrt2-i\)?

I get this to be in the 4th quadrant, but it seems incorrect?...thanx for the help!..

The modulus is 3, and the angle calculates to 30 or 40 something..cannot remember...anycase the actual question is the quadrant...please guys, just help me out with this...i do appreciate!!

juriemagic Sep 13, 2019

#1**+2 **

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.

\(now\\ z=\sqrt{3} \left( \sqrt{\frac{2}{3}}+\frac{-1}{\sqrt{3}} i \; \right)\\ z=r(cos\theta +isin\theta)\\ cos\theta=\sqrt\frac{2}{3}\qquad sin\theta = \frac{-1}{\sqrt{3}}\\ 4th \;\;quad\\ \theta=360-acos(\sqrt\frac{2}{3})\\ \theta \approx 360-35.26^\circ\\ \theta \approx 324.74^\circ\\ \)

Melody Sep 13, 2019

#2**+1 **

Melody,

thank you a million times over...I was going through a question paper with memo, with a student, and saw the memo had the answer as 180- 35.26...in other words they had it in the 2nd quadrant. I assured my student that the memo was incorrect, but I just had to make 100% sure myself, therefore I asked the question. You have confirmed the quadrant. Thank you very much Melody, and also, a BIG thank you for your in depth explanations. I do appreciate. Take care.

juriemagic
Sep 13, 2019