## Precalculus: Mathematics for Calculus, 7th Edition

$(2i)^{4}=16$
$(2i)^{4}=(2i)^{2}\times(2i)^{2}=4i^{2}\times4i^{2}$ $i^{2}=i\times i=\sqrt -1\times\sqrt -1=-1$ hence $4i^{2}\times4i^{2}=4(-1)\times4(-1)=-4\times-4=16$ Hence the answer written in the form $a+bi$ where $a$ is the real part and $b$ is the imaginary part is $16$ because $a=16$ and $b=0$.