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I have $-1 + \tan(3)i$ and must find its modulus and its argument. I tried to solve it by myself for hours, and then I looked at the answer, but I am still confused with a part of the solution.

Here is the provided solution: $\begin{align} z &= -1 + \tan(3)i \\ &= -1 + \frac{\sin(3)}{\cos(3)}i \\ &= \frac1{\left|\cos(3)\right|} ( \cos(3) + i(-1)\sin(3)) \\ &= \frac1{\left|\cos(3)\right|} e^{-3i} \\ &= \frac1{\left|\cos(3)\right|} e^{(2\pi-3)i} \end{align}$

I don't understand how we get to $ \frac1{\left|\cos(3)\right|}(\cos(3) + i(-1)\sin(3)) $ How did they get this modulus $1/|\cos(3)|$, and the $-1$ in the imaginary part? How did they reorder the previous expression to obtain this?

I also don't see why they developed the last equality. They put $2\pi-3$ instead of $-3$; OK, it is the same, but what was the aim of a such development?

Thanks!

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    @Sasha : Thank you very much for your help ! I didn't pay attention to cos(3) < 02011-08-23

3 Answers 3

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Let z=−1+tan(3) i. In the complex plane, this would be the point (−1,tan(3))

, which has length

|z|=(−1)2+tan2(3)−−−−−−−−−−−−−√=1+sin2(3)cos2(3)−−−−−−−−−−√=1cos2(3)−−−−−−√cos2(3)+sin2(3)−−−−−−−−−−−−−√=1|cos(3)|

For the last equality, we used sin2(x)+cos2(x)=1

. Now we want to write

z=|z| eωi=|z| (cos(ω)+isin(ω))

for some ω

. It turns out this can be done easily by writing

z=1cos(3)(−cos(3)+isin(3))=1|cos(3)|(cos(−3)+isin(−3))=1|cos(3)|e−3i

Since −3∉[0,2π) they decided to add 2π to the angle, so that it is inside this interval.

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    At least for me, it is difficult to read.2016-02-17
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Let me fill in some of the steps they have jumped over.

$\begin{align} z &= -1 + i\frac{\sin 3}{\cos 3} \\ &= \frac 1 {\cos 3} (-\cos 3 + i \sin 3) \end{align}$ However, $1/\cos 3$ is a negative real number. To make that term positive, we negate both terms: $\begin{align} z &= \frac 1 {-\cos 3} (\cos 3 - i \sin 3) \\ &= \frac 1 {\lvert \cos 3 \rvert} (\cos 3 - i \sin 3). \end{align}$

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    oh yes i got it, because it it a length ! (and length can't be negative...)2011-08-24
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Let $z = -1 + \tan(3) \ i$. In the complex plane, this would be the point $(-1, \tan(3))$, which has length

$|z| = \sqrt{(-1)^2 + \tan^2(3)} = \sqrt{1 + \frac{\sin^2(3)}{\cos^2(3)}} = \sqrt{\frac{1}{\cos^2(3)}} \sqrt{\cos^2(3) + \sin^2(3)} = \frac{1}{|\cos(3)|}$

For the last equality, we used $\sin^2(x) + \cos^2(x) = 1$. Now we want to write

$z = |z| \ e^{\omega i} = |z| \ (\cos(\omega) + i \sin(\omega))$

for some $\omega$. It turns out this can be done easily by writing

$z = \frac{1}{\cos(3)}(-\cos(3) + i \sin(3)) = \frac{1}{|\cos(3)|}(\cos(-3) + i \sin(-3)) = \frac{1}{|\cos(3)|} e^{-3i}$

Since $-3 \notin [0, 2\pi)$ they decided to add $2 \pi$ to the angle, so that it is inside this interval.

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    Yes thank you... I started to mix everything.. Now I am ok with everything :)2011-08-24