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!
Best Answer
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.