help

Remember that (a+b)² = a² + 2ab + b²

^{My hint may be wrong I did this course a long time ago}

Find

$\left|\left(1+2i\right)^8\right|$.

$\begin{array}{|rclrcl|} \hline && \left|\left(1+2i\right)^8\right| \\ &=& \left|\Big(\left(1+2i\right)^2\Big)^4\right| & \qquad \left(1+2i\right)^2 &=& 1+4i+4i^2 \quad | \quad i^2 = -1\\ & & & &=& 1+4i-4 \\ & & & &=& -3+4i \\ &=& \left|\left(-3+4i\right)^4\right| \\ &=& \left|\Big(\left(-3+4i\right)^2\Big)^2\right| & \qquad \left(-3+4i\right)^2 &=& 9-24i+16i^2 \quad | \quad i^2 = -1\\ & & & &=& 9-24i-16i \\ & & & &=& -7-24i \\ &=& \left|\left(-7-24i\right)^2\right| \\ &=& \left|\left(-1 \right)^2\left(7+24i\right)^2\right| \\ &=& \left| \left(7+24i\right)^2\right| & \qquad \left(7+24i\right)^2 &=& 49+336i+576i^2 \quad | \quad i^2 = -1\\ & & & &=& 49+336i-576i \\ & & & &=& -527+336i \\ &=& \left|-527+336i\right| \\ &=&\sqrt{\left(-527 \right)^2 + 336^2} \\ &=&\sqrt{390625} \\ &=&\mathbf{625} \\ \hline \end{array}$

$\mathbf{\left|\left(1+2i\right)^8\right| = 625}$

Is the following an acceptable way to do it?

$\phantom{=\quad}\left|\left(1+2i\right)^8\right|\\~\\ {=\quad}\left(\left|1+2i\right|\right)^8\\~\\ {=\quad}\left(\sqrt{1^2+2^2}\right)^8\\~\\ {=\quad}\left(\sqrt{5}\right)^8\\~\\ {=\quad}5^{\frac82}\\~\\ {=\quad}5^4\\~\\ {=\quad}625$_

$\phantom{=}|(1 + 2i)^8|\\ = |1 + 2i|^8\\ = \left(\sqrt 5\right)^8\\ = 625$