Let and be the roots of the equation x^2+(2i-1)=0.

Question

Let $\alpha$ and $\beta$ be the roots of the equation $x^2+(2i-1)=0$. Then the value of $|\alpha^8+\beta^8|$ is equal to:

Accepted Answer

$x^2=1-2i$, so $\alpha^2=\beta^2=1-2i$, and $\beta=-\alpha$ (the two square roots are negatives of each other), so $\alpha^8=\beta^8$.

$$\alpha^8=(\alpha^2)^4=(1-2i)^4.$$

$(1-2i)^2=1-4i-4=-3-4i$. $(1-2i)^4=(-3-4i)^2=9+24i-16=-7+24i$.

$$\alpha^8+\beta^8=2(-7+24i)=-14+48i.$$

$$|\alpha^8+\beta^8|=2|{-7+24i}|=2\sqrt{49+576}=2\sqrt{625}=2(25)=50.$$

Solutions · 1