Inverse z-Transformation and impulse response

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I have a question regarding a pole zero plot and I want to compute the impulse response with the inverse z-Transform.

From the pole zero plot I could extract:

$$H(z) = \frac{z+1}{(z-0.6i)(z+0.6i)}$$

How can I get $h[n]$?

If I use partial fraction expansion I get something complex for $A_1$ and $A_2$ and that doesn't make sense, at least to me.

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Partial fractions will work and the end result will be real valued:

$$H(z)=\frac{Az}{z-0.6i}+\frac{A^*z}{z+0.6i}\tag{1}$$

where $A=0.5-i\cdot 5/6$, and $^*$ denotes complex conjugation. The inverse $\mathcal{Z}$-transform of $(1)$ is

$$\begin{align}h[n]&=\left[A(0.6i)^n+A^*(-0.6i)^n\right]u[n]\\&=\left[A(0.6)^ne^{in\pi/2}+A^*(0.6)^ne^{-in\pi/2}\right]u[n]\\&=2\,\text{Re}\left\{A(0.6)^ne^{in\pi/2}\right\}u[n]\\&=2(0.6)^n|A|\cos\left(\frac{n\pi}{2}+\arg(A)\right)\tag{2}\end{align}$$

which is real-valued, as it should be.