Answer to Question #117183 in Algebra for Edward

Question

Find the modulus and the principal argument of each of the given complex numbers.
(a) 3 − 4i, (b) −2 + i, (c) 1
1 + i
√
3
, (d) 7 − i
−4 − 3i
,
(e) 5(cos π/3 + isin π/3), (f) cos 2π/3 − sin 2π/3.

Answer 1

We will use the notation $z = a+ib$ for given complex number. So, modulus of $z$ is

$|z| = \sqrt{a^2+b^2}$ and principal argument of $z$ is $arg(z) = tan^{-1}(\frac{b}{a}) \in (-\pi,\pi).$

a) Given $z = 3-4i$

Hence, $a = 3, b = -4$.

$\implies |z| = \sqrt{3^2+(-4)^2} = 5$

And, since z lies in fourth quadrant so $arg(z) = - tan^{-1} (\frac{4}{3})$.

b) Given $z = -2+i$ .

So, $|z| = \sqrt{2^2+1^2} = \sqrt{5}$

and z lies in 2nd quadrants so $arg(z) = \pi - tan^{-1}(\frac{1}{2})$.

c) Given $z = 1+i\sqrt{3}$ .

Hence, $|z| = \sqrt{1^2+3} = \sqrt{4} = 2$

and z lies in first quadrant so $arg(z) = tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$

d) Given $z = -4-3i$.

Hence, $|z| = \sqrt{4^2+3^2} = \sqrt{25} = 5$

and z lies in 3rd quadrants, so $arg(z) = \pi+tan^{-1}(\frac{3}{4})$.

e) Given $z = 5(cos (π/3) + isin( π/3))$

Hence, $|z| = \sqrt{5^2(cos^2(\pi/3)+sin^2(\pi/3))} = 5$

And $arg(z) = tan^{-1}(\frac{5sin(\pi/3)}{5cos(\pi/3)}) = tan^{-1} (tan(\pi/3)) = \pi/3$ .

f) Given $z = cos(2\pi/3) - sin(2\pi/3) = -cos(\pi/3)-sin(\pi/3) = - \frac{1}{2} - \frac{\sqrt{3}}{2}$.

So, $|z| = \sqrt{( cos(2\pi/3) - sin(2\pi/3))^2} = cos(2\pi/3) - sin(2\pi/3) = - (\frac{1+\sqrt{3}}{2})$

and given z lies on negative $x-$axis, hence $arg(z) = tan^{-1}(0) = -\pi$ .