If z=(p+1)^4 and p is real, find the values for p

Question

Given that z=(p+i)^4, where p is real, find the values of p for which..

A) z is real B) z is a real multiple of i

I attempted doing this by opening up the brackets, which then lead to..

z= p^4 + 4p^3i - 6p^2 -4pi + 1

After this I got lost. In A) I tried denotating: z=a (since z is real). I thought I had to use the quadratic formula, since the equation above also equals to (p^2 + 2pi -1 )(p^2 + 2pi -1). That resulted in -i, which isn't the answer.

Answer for A) should be: 0, ±1 and for B) : ±(1+√2), ±(1-√2)

Answer 1

Hint $z^4=\operatorname{Re}+i\operatorname{Im}=(p^4-6p^2+1)+i(4p^3-4p)$. If $z^4$ is real the imaginary part is $0$. If $z^4$ is a multiple of $i$ then real part is 0.

Answer 2

A complex number is real iff $\bar z=z$, that is $z/\bar z=1$, provided $z\neq0$. In our case this gives $$\left(\frac{p+i}{p-i}\right)^4=1\iff\frac{p+i}{p-i}\in\{\pm1,\pm i\}.$$