Simplify your answer completely: $5i(1 + i)^2$ Simplify your answer completely: $5i(1 + i)^2$
I know the answer is $-10$, but I don't know how to get it. Things I tried:
foiling $(1+i)^2$ and then distributing $5i$.
distributing $5i$ into $(1+i)$ and then foiling
 A: Since 
$$(a+b)^2=a^2+2ab+b^2,$$ one has$$5i(1+i)^2=5i(1^2+2\cdot 1\cdot i+i^2)=5i(1+2i-1)=5i\cdot 2i=-10.$$
A: Foiling first is the correct path to take, so once expanded, you can then multiply by $5i$. To foil, remember: $$(a + b)^2 = a^2 + 2ab + b^2$$ That works for real and non-real numbers, $a, b$.
$$5i[(1+i)^2]=5i(1+2i+i^2)=5i(1+2i -1) =5i\cdot 2i = 10\cdot i^2 = -10$$
A: $$5i(1 + i)^2=$$
$$5i(1+i)(1+i)=$$
$$(\left|5i\right|e^{arg(5i)i})(\left|1+i\right|e^{arg(1+i)i})(\left|1+i\right|e^{arg(1+i)i})=$$
$$\left(5e^{\frac{1}{2}\pi i}\right)\left(\sqrt{2}e^{\frac{\pi}{4}i}\right)\left(\sqrt{2}e^{\frac{\pi}{4}i}\right)=$$
$$\left(5e^{\frac{1}{2}\pi i}\right)\left(2e^{\frac{1}{2}\pi i}\right)=$$
$$10e^{\pi i}=$$
$$10\left(\cos(\pi)+\sin(\pi)i\right)=$$
$$10(-1+0i)=$$
$$-10$$
A: $\arg (1+i) = \frac{\pi}{4}$ and $\arg (i) = \frac{\pi}{2}$, therefore $\arg(5i(1 + i)^2) = \arg(i) + 2\arg(1+i) =  \pi$.
Then $|1+i| = \sqrt{2}$, $|5i|=5$, so $|5i(1+i)^2|=|5i|\,|1+i|^2 = 10$.
Together you will have $5i(1 + i)^2 = 10\angle \pi = -10$.
I think in this way it is easier to visualize.
