First of all $i^{27}=-i$ but I think that you just had a finger error. Then you have that $z=\frac{r^{n+1}-1}{r-1}$ with $r=2i$, then \begin{align*} z=\frac{-2^{27}i-1}{2i-1}&= \frac{-2^{27}i-1}{2i-1}\cdot \frac{2i+1}{2i+1}\\ &=\frac{2^{28}-1-(2^{27}+2)i}{-3} \end{align*} So, $Re(z)=\frac{2^{28}-1}{-3}$, $Im(z)=\frac{(2^{27}+2)i}{3}$ and \begin{align*} |z|&=\sqrt{\left(\frac{2^{28}-1}{-3}\right)^2+\left(\frac{(2^{27}+2)i}{3}\right)^2}\\ &=\sqrt{\frac{2^{56}+2^{54}+5}{9}} \end{align*}

First of all $i^{27}=-i$ but I think that you just had a finger error. Then you have that $z=\frac{r^{n+1}-1}{r-1}$ with $r=2i$, then \begin{align*} z=\frac{-2^{27}i-1}{2i-1}&= \frac{-2^{27}i-1}{2i-1}\cdot \frac{2i+1}{2i+1}\\ &=\frac{2^{28}-1-(2^{27}+2)i}{-5} \end{align*} So, $Re(z)=\frac{2^{28}-1}{-5}$, $Im(z)=\frac{(2^{27}+2)i}{5}$ and \begin{align*} |z|&=\sqrt{\left(\frac{2^{28}-1}{-3}\right)^2+\left(\frac{(2^{27}+2)i}{5}\right)^2}\\ &=\sqrt{\frac{2^{56}+2^{54}+5}{25}} \end{align*}