Given a complex number, convert it to the form $a + bi$ So I'm trying to convert a complex number to the form of $a + bi$. The complex number in question is $$\bbox[5px,border:2px solid #C0A000]{\large 2e^{\frac{-3\pi}{4i}+\ln(3)}}$$ I'm not quite sure how to tackle this to be honest, I would appreciate some help in trying to understand how to convert these expressions into that form.
Thanks in advance!
 A: The first thing you can do is factor the expression into $2e^{\ln(3)} \times e^{-3\pi/4i}$.
Then, we can use Euler's formula: $e^{ix} = \cos x + i \sin x$.
If you apply that formula to the expression, you will be able to get your answer with a little bit of arithmetic and simplification.
A: there are two standard forms for complex numbers.
Rectangular form $z =a+bi$ where $a = Re(z)$ and $b = Im(z)$.
And, if $z \ne 0$ polar form  $z = r e^{\theta i}$ where $r = |z|$ and $\theta =\arg(z)$.
$e^{\theta i}$ is defined to be $e^{\theta i} = \cos \theta + i \sin \theta$.
So there is a conversion between these two forms.
If $z = r e^{\theta i} = r\cos \theta + (r \sin \theta) i$.
And if $z = a + bi= \sqrt{a^2 + b^2} (\cos (\arctan \frac ba) + \sin (\arctan \frac ba)i) = \sqrt{a^2 + b^2} e^{\arctan \frac ba i}$.
....
Now the only trick this problem has done is it has combined then $r=|z|$ and $e^{\theta i}$ by replacing $r$ with $e^{\ln r}$ and combining $r e^{\theta i} = e^{\ln r} e^{\theta i} = e^{\theta i + \ln r}$ to get it as a single exponent.
.....
So $2e^{-3\pi/4i+ln(3)}=$
$2e^{-\frac {3\pi}4i}e^{\ln 3} =$
$6e^{-\frac {3pi}4i} = 6(\cos(-\frac {3\pi}4)+ i\sin (-\frac {3\pi} 4))=$
$6(-\frac {\sqrt 2}2 + \frac {\sqrt 2}2i) =$
$-3\sqrt 2 +3\sqrt 2 i$.
That's it.
A: First, convert it into $re^{i \theta}$. Then we have $a = r \cos \theta$ and $b = r \sin \theta$. In your case, we have $r = 2e^{\ln 3} = 6$ and $\theta = -3 \pi / 4$.
