# Complex number + trigo : $-1 + \tan(3)i$ , find modulus and argument

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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Given the solution, are you sure you don't have $-1 + i\tan(3)$ instead of $-1 + \tan(3)$? –  Rahul Aug 23 '11 at 20:58
They notice that $\cos(3) < 0$, and use $\frac{\sin (3)}{\cos (3)} = \frac{\sin( 3) \operatorname{sgn}(\cos(3))}{ \vert \cos (3) \vert} = \frac{(-1) \sin( 3)}{ \vert \cos (3) \vert}$. –  Sasha Aug 23 '11 at 21:00
@Rahul: you are right ! I mistyped –  0alpha0 Aug 23 '11 at 21:02
@Sasha : Thank you very much for your help ! I didn't pay attention to cos(3) < 0 –  0alpha0 Aug 23 '11 at 21:38

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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I misunderstood abs and module notation for |cos(3)| , I thought it was module... Thank you for this wonderful answer ! –  0alpha0 Aug 23 '11 at 21:27
@Jonathan: But the modulus and absolute value of a real number are the same thing... –  Ｊ. Ｍ. Aug 24 '11 at 3:45
Yes thank you... I started to mix everything.. Now I am ok with everything :) –  0alpha0 Aug 24 '11 at 8:32

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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why do we want it to be positive ? –  0alpha0 Aug 23 '11 at 21:28
@Jon: Because moduli are intended to be positive. –  Ｊ. Ｍ. Aug 24 '11 at 2:50
oh yes i got it, because it it a length ! (and length can't be negative...) –  0alpha0 Aug 24 '11 at 8:33