The given question is very simple. I am simply told to draw phase portraits of vector field $z'=iz$. However, I am having a very hard time to just picture this in my head. I am not quite getting it and would like help on how would I approach to draw this.
I know that if we have $z=x+yi$, the matrix representation of $iz$ would be $\begin{bmatrix}-y & x\\-x & -y\end{bmatrix}$
I wonder if this matrix has anything to do with this question. Help would be appreciated it. Thanks.
On the one hand you can say
for any constant $k,z' = kz$ implies $z = e^{kt}$
and $i$ is just a constant.
$z = z_0 e^{it}\\ z_0 = \rho e^{i\theta}\\ z = \rho e^{i(\theta + t)}$
If you want to use matrix notation.
$z = x+iy = \begin{bmatrix} x\\y \end{bmatrix}$
$z'= iz = ix - y = \begin{bmatrix} 0&-1\\1&0 \end{bmatrix}z$
But then you still have:
$z = e^{\begin{bmatrix} 0&-1\\1&0 \end{bmatrix}t}z_0$
Now you plug $\begin{bmatrix} 0&-1\\1&0 \end{bmatrix}t$ into the Taylor series:
$e^{At} = \sum \frac {(At)^n}{n!}$
$A^2 = -I\\ A^3 = -A\\ A^4 = I$
$e^{At} = \sum \frac {(-1)^nIt^{2n}}{2n!} + \frac {(-1)^nAt^{2n+1}}{(2n+1)!}$