Finding argument of complex number and conversion into polar form How do I find the argument of a complex number, for example $z = 3 + 4i$?
I know the polar form of $z$ is $r(\cos\theta + i\sin\theta)$ where $r$ is the modulus of $z$ ($\sqrt{3^2+4^2}$) which would leave me with $z = 5(\cos\theta+i\sin\theta)$, I'm just unsure how to deduce $\theta$ here.
The textbook says two things: $\theta = \arg(z)$ and $\tan\theta = b/a$, I'm not sure how these help.
 A: Hint: irrespective of $|z|$, you will get $\tan(\theta)=\frac{4}{3}$. So, taking the inverse will give you $\theta$ in radians.
A: Your textbook says that if you have $z = a+ bi$ then $\tan \theta = \frac{b}{a}$ so that $$\theta = \arctan \frac{b}{a}$$
In your case, this is $\theta = \arctan \frac{4}{3}$ for $z = 3+ 4i$. Although, really, the convention for the argument makes this not so straightforward. The Wikipedia article explains the computation of the argument well.
A: It is true that $\tan\theta=\frac{b}{a}$, but you have to be careful because $\tan\theta=\tan(\theta+\pi)$. If $a$ and $b$ are both negative, $\arctan(b/a)$ will not give you the right answer, you have to do $\arctan(b/a)+\pi$ (since $b/a=(-b)/(-a)$). You also have to be careful if one of $a$ or $b$ is negative.
A: For the first quadrant in the complex plane we know that the argument is given by:
$$\arg\left(z\right)=\tan^{-1}\frac{\Im(z)}{\Re(z)}\Longrightarrow \arg(a+bi)=\tan^{-1}\left(\frac{b}{a}\right)$$
And we know that:
$$a+bi=|a+bi|e^{\arg(a+bi)i}=|a+bi|(\cos(\arg(a+bi))+\sin(\arg(a+bi))i)$$

$$3+4i=|3+4i|e^{\arg(3+4i)i}=|3+4i|(\cos(\arg(3+4i))+\sin(\arg(3+4i))i)=$$
$$\sqrt{3^2+4^2}e^{\tan^{-1}\left(\frac{4}{3}\right)i}=\sqrt{25}e^{\tan^{-1}\left(\frac{4}{3}\right)i}=5e^{\tan^{-1}\left(\frac{4}{3}\right)i}$$
So the three forms are:
$$3+4i=$$
$$5e^{\tan^{-1}\left(\frac{4}{3}\right)i}=$$
$$5\left(\cos\left(\tan^{-1}\left(\frac{4}{3}\right)\right)+\sin\left(\tan^{-1}\left(\frac{4}{3}\right)\right)i\right)$$
A: If $ Z =a+ib$ is a complex number 
$|Z|=\sqrt{a^2+b^2}$
$\theta=argZ$
if Z lies in $Q_{1} then \theta=\tan^{-1}(\frac{b}{a})$
if Z lies in $Q_{2}then \theta=\pi-tan^{-1}(\frac{b}{a})$
if Z lies in $Q_{3} then \theta=-\pi+tan^{-1}(\frac{b}{a})$
if Z lies in $Q_{4} then \theta=-tan^{-1}(\frac{b}{a})$
A: Question: Let $z=a+ib.$ Why does the following formula not hold in general?
$$\mathrm{Arg}(z)=\arctan\left(\frac ba\right)$$
Answer: Suppose that $b<0.$
Then $Z \text{ lies in quadrant }2\;\text{or}\;3.$
So $\;\mathrm{Arg}(z)>\frac{\pi}2\quad\text{or}\quad\mathrm{Arg}(z)<-\frac{\pi}2.$
But since $-\frac\pi2<\arctan\left(\frac ba\right)<\frac\pi2,\quad\mathrm{Arg}(z)$ cannot possibly equal $\arctan\left(\frac ba\right).$

There are several compact formulae for $\mathrm{Arg}(z)$, one of which is $$\mathrm{Arg}\left(a+ib\right)=\begin{cases}
 \quad\arccos \left(\frac a{\sqrt{a^2+b^2}}\right) \text{ if } b\geq0;\\
 -\arccos \left(\frac a{\sqrt{a^2+b^2}}\right) \text{ if } b<0.\end{cases}$$
However, it is easiest to just use an Argand diagram:

*

*locate the angle between $OZ$ and its projection on the real axis;

*evaluate it by taking $\arctan$ of the corresponding ratio of
lengths;

*take the adjacent supplementary angle iff Re$(z)<0;$

*assign a negative sign iff the resultant angle is directed clockwise from the positive real axis;

*the desired principal argument is thus obtained.


For example, $$\mathrm{Arg}\left(-2-3i\right)\\=-\left[\pi-\arctan\left(\frac32\right)\right]\\=\arctan\left(\frac32\right)-\pi.$$
