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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.

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

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.

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