Multiplying complex vector by a complex number How can I geometrically interpret multiplying a complex vector by a complex number. Let's say I have a vector $(a, b)$ in a complex vector space. Now, let's say I multiply this vector by $re^{i\theta}$. Is the resultant vector rotated and stretched like in real vector spaces?
 A: The short answer is (I think) "no". If you imagine your vector space as having four real dimensions then each pair of real components (first and second, third and fourth) is a complex plain in which the corresponding component is rotated and stretched.
If a better geometric answer shows up I'll delete this one.
A: If your vector space is over $\mathbb{C}$ then multiplication by a complex scalar keeps you in the same one-dimensional subspace spanned by your original vector. Then I think it is difficult to talk of rotation. However, if you consider your vector space over $\mathbb{R}$ then your transformation with $r=1$ preserves Euclidean norm and in principle takes you out of the span of your original vector so this would correspond to a rotation. Different values of $r>0$ then correspond to stretching or shrinking.
A: What happens to a complex vector when multiplied by a complex scalar is easier to understand if you think of complex vectors as ellipses instead of arrows.
Short answer
The complex vector is scaled but not rotated.
Some background
Complex vectors have a one-to-one correspondence with real time-harmonic field vectors.
Let ${\bf F}(t)$ be any real vector function of time $t$ that satisfies the differential equation
$\frac{d^2}{dt^2}{\bf F}(t) + \omega^2 {\bf F}(t) = 0$
A general solution can be expressed with two real vectors ${\bf F_1}$ and ${\bf F_2}$ in the form
${\bf F}(t) = {\bf F_1} \mathrm{cos}\omega t + {\bf F_2} \mathrm{sin}\omega t$
Let ${\bf f} = {\bf f}_{re} + i {\bf f}_{im}$,
where ${\bf f}_{re}$ and ${\bf f}_{im}$ are both real vectors.
With two (each other's inverse) mappings we see the correspondence ${\bf F}(t) \leftrightarrow {\bf f}$:
${\bf f} \rightarrow {\bf F}(t): {\bf F}(t) = \mathrm{Re}\{{\bf f} e^{i \omega t}\} = {\bf f}_{re} \mathrm{cos} \omega t - {\bf f}_{im}\mathrm{sin} \omega t$
${\bf F}(t) \rightarrow {\bf f}: {\bf f} = {\bf F}(0) - i {\bf F}(\pi / 2 \omega) = {\bf F_1} - i {\bf F_2}$
The geometrical interpretation
The time-harmonic real vector ${\bf F}(t)$ traces an ellipse as time varies. The real part of the corresponding complex vector ${\bf f}$, ${\bf f}_{re}$, defines the time-harmonic vector at time $t = 0$, as can be seen from above.
Thus, while real vectors can be interpreted as arrows in space, complex vectors can be interpreted as ellipses in space. As the phase of the complex vector changes, the 'instantaneous' arrow of the time-harmonic counterpart traces an ellipse.
Now what happens when you multiply a complex vector ${\bf f}$ with a complex scalar $\lambda e^{i \phi}$? Using the first mapping above
${\bf G}(t) = \mathrm{Re}\{\lambda e^{i \phi} {\bf f} e^{i \omega t}\} = \mathrm{Re}\{\lambda ({\bf f}_{re} + i{\bf f}_{im})e^{i(\omega t + \phi)}\} = \lambda ({\bf f}_{re} \mathrm{cos} (\omega (t + \frac{\phi}{\omega})) - {\bf f}_{im} \mathrm{sin}(\omega(t + \frac{\phi}{\omega}))) = \lambda {\bf F}(t + \frac{\phi}{\omega})$
Thus, multiplying the complex vector representation with a complex scalar is the same as scaling the time-harmonic real vector and changing the direction of its origin (i.e. direction at $t = 0$). In other words, the ellipse defined by the complex vector is scaled but not rotated.
A much more thorough explanation can be found from chapter 1 of Methods for Electromagnetic Field Analysis, on which I based this answer.
