# How To Compute Args and Mods of Complex Numbers That Satisfy Inequalities

A complex number $z$ satisfies the inequality

$$|z + 2 - (2\sqrt{3})i|\le 2$$

Find the least possible value of $|z|$ and the greatest possible value of $argz$

the answers given in the text book is $2$ and $\frac{5\pi}{6}$ respectively.

Even when I plot the equation on an argand diagram I am still struggling to isolate $|z|$ and understand how they are getting those answers.

Please suggest the best way to compute these values as I have been stuck on this question for two days?

PS. this is not homework, but prep study before I start my degree in computer science in the UK

Thanks in advance

## 3 Answers

Take a sheet of graph paper and pick out where $\;-2+2\sqrt3\,i$ would be. Consider it's angle and recognize that it is a distance of 4 from the origin. Then draw a circle of radius two around that point. All the points inside or on the curve satisfy the inequality. $\;|z|$ is the distance from the origin. From there it is just considering what makes sense graphically.

• thanks for explaining in such simple terms, I would love to share the answer between you and AlexR, however I will award it to you for your succinctness and brevity – user866190 Aug 29 '13 at 10:52

Hint:

The equation $\;|z-(-2+2\sqrt3\,i)|\le2\;$ describes a disk of radius $\,2\;$ and centered on $\,-2+2\sqrt3\,i\;$ , and thus the maximal modulus and argument are (further hint: argument is defined only up to multiples of...)

• thanks for the hint.. but why is the radius the square root of 2..? – user866190 Aug 29 '13 at 10:32
• $|z-y|\leq 2$ gives a radius of $2$, not $\sqrt{2}$. – AlexR Aug 29 '13 at 10:36
• Indeed @AlexR, thanks for the catch. – DonAntonio Aug 29 '13 at 10:38
• Read the edited answer, @user866190 – DonAntonio Aug 29 '13 at 10:39

$$|z- (-2+2\sqrt{3} i)| \leq 2 \Rightarrow z \in \overline{B_2(-2+2\sqrt{3}i)}$$ So $|z| \geq |-2+2\sqrt{3}i| - 2 = 4-2 = 2$.

Can you figure out $\arg(z) \in A \subset [0, 2\pi)$ yourself?

• thanks for the help, confused with your set notation B and A. please clarify. – user866190 Aug 29 '13 at 10:45
• @user866190 $$B_\delta(y) = \{z: |z-y| < \delta\}$$ The "ball" around $y$ with radius $\delta$, $A$ is a general set – AlexR Aug 29 '13 at 10:55
• @user866190 Hint: $A$ will be a closed Interval, if you identify $0$ with $2\pi$, i.e. $$A = [0, a_1] \cup [a_2, 2\pi)$$ or $$A = [a_2, a_1]$$ – AlexR Aug 29 '13 at 13:03
• Please modify "$\arg(z) \subset A$". – Did Aug 29 '13 at 19:56
• @Did done, thanks. – AlexR Aug 31 '13 at 16:25