# [SOLVED]Elementary Complex Number Problems

#### nacho

##### Active member
1.
$|\frac{i(2+i)^3}{(1-i)^2}|$

Is there any way to complete this without expanding the numerator?

2. what is the argument of $-2\sqrt{3}-2i$
I got $r=4$

then
$\cos\theta_1$ $= \frac{-2}{\sqrt{3}{4}}$ and $-2=4\sin\theta_2$
$\theta_1 = \pi - \frac{\pi}{6} = 5\frac{\pi}{6}$ and
$\theta_2 = \frac{-\pi}{6}$
so
Arg = $4cis(\frac{-\pi}{3})$ which is wrong according to my solutions and it should be $\frac{-5\pi}{6}$

where did i go wrong?

2. what is the method of finding the argument of -1/2.

so $z = -\frac{1}{2}$
and $r = \frac{1}{2}$

to solve for theta, i always get confused here.

i let
$-\frac{1}{2} = \frac{1}{2}\cos\theta_{1}$ and $-\frac{1}{2}=\frac{1}{2}\sin\theta_{2}$ and solve

usually my $\theta_{2}$ ends up being wrong due to some error i make in the range. What would I do from here, being as meticulous and thorough in my working as possible?
in this example, i made no mistake.
i will edit this section with a question i get wrong, of similar fashion.
Thanks.

Last edited:

#### chisigma

##### Well-known member
Re: elementary problems

2. what is the method of finding the argument of -1/2.
In my opinion the 'good definition' of $\displaystyle \text{arg}\ z$ is the following...

$\displaystyle \text{arg}\ z = \mathcal{Im} (\ln z)\ (1)$

On the basis of (1) is...

ln (- 1/2) - Wolfram|Alpha

$\displaystyle \text{arg}\ (- \frac{1}{2}) = \pi\ (2)$

Kind regards

$\chi$ $\sigma$

#### nacho

##### Active member
Re: elementary problems

how would i find the argument of
$(1-i)(-\sqrt{3} + i)$
I get $r = 2\sqrt{2}$

but when solving for theta i get stuck because it gives:

$\frac{\sqrt{3}+1}{2\sqrt{2}} = \cos\theta$ which i cannot solve.

#### chisigma

##### Well-known member
Re: elementary problems

how would i find the argument of
$(1-i)(-\sqrt{3} + i)$
I get $r = 2\sqrt{2}$

but when solving for theta i get stuck because it gives:

$\frac{\sqrt{3}+1}{2\sqrt{2}} = \cos\theta$ which i cannot solve.
Also in this case 'Monster Wolfram' works excellently!...

ln [(1 - i)/(i - sqrt(3))] - Wolfram|Alpha

$\displaystyle \text{Im}\ (\ln \frac{1-i}{i - \sqrt{3}}) = \frac{11}{12}\ \pi\ (1)$

Kind regards

$\chi$ $\sigma$

#### nacho

##### Active member
we are not expected to solve these by hand?

#### Prove It

##### Well-known member
MHB Math Helper
1.
$|\frac{i(2+i)^3}{(1-i)^2}|$

Is there any way to complete this without expanding the numerator?
Yes, convert to polars.

#### Plato

##### Well-known member
MHB Math Helper
1.
$|\frac{i(2+i)^3}{(1-i)^2}|$
Is there any way to complete this without expanding the numerator?
$$\left| {\frac{{i{{(2 + i)}^3}}}{{{{(1 - i)}^2}}}} \right| = \frac{{|i||2 + i{|^3}}}{{|1 - i{|^2}}}$$

#### nacho

##### Active member
$$\left| {\frac{{i{{(2 + i)}^3}}}{{{{(1 - i)}^2}}}} \right| = \frac{{|i||2 + i{|^3}}}{{|1 - i{|^2}}}$$
I just wasn't sure how to proceed with the $|(2+i)|^3$ term.
would i need expand the entire term or?

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Re: elementary problems

In my opinion the 'good definition' of $\displaystyle \text{arg}\ z$ is the following...

$\displaystyle \text{arg}\ z = \mathcal{Im} (\ln z)\ (1)$

On the basis of (1) is...

ln (- 1/2) - Wolfram|Alpha

$\displaystyle \text{arg}\ (- \frac{1}{2}) = \pi\ (2)$

Kind regards

$\chi$ $\sigma$
I think what is meant by argument is finding the angle $$\displaystyle \theta$$ of the complex number vector with the x-axis.

#### Plato

##### Well-known member
MHB Math Helper
I just wasn't sure how to proceed with the $|(2+i)|^3$ term.
would i need expand the entire term or?
$$\displaystyle |2+i|=\sqrt{5}$$ so $$\displaystyle |2+i|^3=\sqrt{5}^3$$