$$(1+i)(e^{(1+i)\phi})$$
I need to express this in both polar and rectangular form, but the difficult part is that extra $i$ above the $e$. Also, what am I supposed to make of $\phi$? We normally use $\theta$ to represent the argument of a complex number, so would this just be another notation for the angle?
Would the final answer in rectangular form not have a $\phi$ in it?
Firstly, $\phi$ is used to denote an angle, just as $\theta$ commonly is. It doesn't matter what variable we use. Although, $\phi$ or $\theta$ are commonly used we could just as easily use $x$ or $y$.
Now on to expanding your expression:
$$(1+i)(e^{(1+i)\phi}) = e^{(1+i)\phi}+ie^{(1+i)\phi}$$
Note that $e^{(1+i)\phi} = e^\phi e^{i\phi} = e^\phi [\cos(\phi)+i\sin(\phi) ]$
Edit:
Here is some more info for you to consider. $$(1+i)(e^{(1+i)\phi}) = e^{(1+i)\phi}+ie^{(1+i)\phi}$$ $$ = e^\phi(\cos(\phi)+i\sin(\phi))+e^\phi(-\sin(\phi)+i\cos(\phi))$$
Noting that $-\sin(\phi)= \cos(\phi+\frac\pi2)$ and $\cos(\phi)=\sin(\phi +\frac\pi2)$ we can further simplify the above expression. So,
$$e^\phi(\cos(\phi)+i\sin(\phi))+e^\phi(-\sin(\phi)+i\cos(\phi))= $$
$$e^\phi(\cos(\phi)+i\sin(\phi))+e^\phi(\cos(\phi+\frac\pi2)+i\sin(\phi+\frac\pi2))$$
Written in this form you can see that your original expression is the sum of two complex numbers with a phase angle difference of $\frac\pi2$, that both have the same magnitude, namely $e^\phi$. To actually write this in terms of cartesian coordinates in any easy way, I believe (I may be wrong) that actual values of $\phi$ would be helpful. Otherwise, refer to this blerb from wikipedia:
I hope this is not too confusing.
