Confusion in converting the polar form back to the rectangular form For example there's a complex number: 
$$-1+4i$$
Now if I find the argument it will be 
$$\arctan(-4)$$
But if I use the argument to convert the polar form back to rectangular form using
$$R\cos\theta$$ 
It gives $1$ as the real part. However if I use 
$$\arctan(-4)+\pi$$ 
I get the correct form.
So my question is how can I know when to add $\pi$ to argument when converting back to rectangular form? 
 A: The image of $\arctan$ gives $2$ quadrants only.
A more practicable formula is 
$$\arg z=2\tan^{-1} \frac{y}{x+\sqrt{x^2+y^2}}$$ and some computer languages using $$\operatorname{atan2}(y,x)$$
A: You seem to have an error in your initial calculation. Note that
$$R=\sqrt{(-1)^2+4^2}=\sqrt{17}$$
$$\phi=\arg\left(-1+4i\right)=\pi-\arctan\left(4\right)$$
Therefore
$$Re^{i\phi}=\sqrt{17}e^{i(\pi-\arctan(4))}$$
$$=\sqrt{17}\left(-\cos(\arctan(4))+i\sin(\arctan(4))\right)$$
$$=\sqrt{17}\left(-\frac{1}{\sqrt{17}}+i\frac{4}{\sqrt{17}}\right)$$
$$=-1+4i$$
Check out this link to see how to properly evaluate the argument of a point in the complex plane. 
A: The basic function $\dot{\mathbb R}^2\to{\mathbb R}/(2\pi)$, resp., $\dot{\mathbb C}\to{\mathbb R}/(2\pi)$, giving the polar angle of a point ${\bf z}=(x,y)\ne{\bf 0}$, resp., a complex number $z=x+i y\ne0$, is the argument function ${\rm arg}$. If we exclude points on the negative $x$-axis  we can select among the "infinitely many values" of ${\rm arg}$ the so called principal value ${\rm Arg}$ of the argument, which then is a number in the interval $\ ]{-\pi},\pi[\ $. One has ${\rm Arg}(z)=0$ on the positive real axis, as well as ${\rm Arg}(\bar z)=-{\rm Arg}(z)$. If $x>0$, and only then, the equation
$${\rm Arg}(x,y)={\rm Arg}(x+iy)=\arctan{y\over x}$$
is valid.
