MHB Elementary Complex Number Problems

[nacho-man]

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.

 

[chisigma]

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-man]

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]

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-man]

we are not expected to solve these by hand?

 

[Prove It]

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

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

Yes, convert to polars.

 

[Plato2]

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-man]

\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?

 

[alyafey22]

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 $$\theta$$ of the complex number vector with the x-axis.

 

[Plato2]

I just wasn't sure how to proceed with the $|(2+i)|^3$ term.
would i need expand the entire term or?

$$|2+i|=\sqrt{5}$$ so $$|2+i|^3=\sqrt{5}^3$$