How do you divide a complex number with an exponent term? Ok, so basically I have this:
$$ \frac{3+4i}{5e^{-3i}} $$
So basically, I converted the numerator into polar form and then converted it to exponent form using Euler's formula, but I can have two possible solutions.
I can have $5e^{0.972i}$ (radian) and $5e^{53.13i}$ (angle).
So my real question is, the exponent, does it have to be in radian or angle?
I ended up with $e^{3.927i}$.
 A: Put the numerator also in e-power form. r = 5 and the argument follows from arctan(4/3)
Now you have an e-power in the Numerator and an e-power in the Denominator. Divide the r-values, subtract the arguments. Don't expect a nice looking answer; arctan(4/3) does not come out good
A: the exponent is always is radians. converting into polar form would give you $$r e^{i \theta}$$ where $$r=\sqrt{3^2 + 4^2}=5$$ and $$\theta=\tan^{-1}(\frac43)$$ or $\arctan (\frac43)$ which gives $$\theta=0.927$$ so your fraction becomes $$\frac{5e^{0.927i}}{5e^{-3i}}$$
$$=e^{3.927i}$$
A: A complex number $z=x+iy$ can be written as $z=re^{i\theta}$, where $r=|z|=\sqrt{x^2+y^2}$ is the absolute value of $z$ and $\theta=\arg{z}=\operatorname{atan2}(y,x)$ is the angle between the $x$-axis and $z$ measured counterclockwise and in radians. In this case, we have $r=5$ and $\theta=\arctan\frac{4}{3}$ (since $x>0$, see atan2 for more information), so $3+i4=5e^{i\arctan\frac{4}{3}}$ and
$$\frac{3+i4}{5e^{-i3}}=\frac{5e^{i\arctan\frac{4}{3}}}{5e^{-i3}}=e^{i(\arctan\frac{4}{3}+3)}\approx e^{i3.927}.$$
A: Radians are indeed conventional, and using degrees will lead to trouble once you start messing around with derivatives and integrals. Whenever an angle appears "bare" in a mathematical formula (without, e.g., a degree symbol, indication of gradians, turns, etc.), it is assumed to be in radians.
You can either get both numbers into rectangular form or both into polar form.
With both in polar form, $$\frac{re^{i\alpha}}{qe^{i\beta}} = \frac r q e^{i\alpha} e^{-i\beta} = \frac r q e^{i(\alpha - \beta)}.$$
With both in rectangular form, $$\frac{a+bi}{c+di} = \frac{a+bi}{c+di}\frac{c-di}{c-di} = \frac{(a+bi)(c-di)}{c^2+d^2} = \frac{ac+bd}{c^2+d^2}+\frac{bc-ad}{c^2+d^2}i.$$
