Let $\displaystyle Z_{n+1}=Z_{n}^{2}+2i$ and $\displaystyle Z_0=0.5+02.i$. Find $\displaystyle Z^{10}$ value and hence deduce the $\displaystyle Z_n$ when $\displaystyle n \longrightarrow \alpha$

here my calculations are

$\displaystyle Z_1=(0.5+0.2i)^2+2i=(0.2+2.2i)$

$\displaystyle Z_2=(0.2+2.2i)^2+2i=(-4.8+2.88i)$

$\displaystyle Z_3=(-4.88+2.88i)^2+2i=(14.7456-25.648i)$

.......

Is that above answer is true?

But I saw in other way like $\displaystyle Z_{n+1}=Z_{n}^{2} + \lambda $ and then $\displaystyle Z^2=r^2e^{i2\theta}$ bt I couldn't understand this..