Question

-64=(2cos(pi/2)+2i sin(pi/2))^n..... n=??​

Answer 1

Answer:

The real solution is $n=6$.

Step-by-step explanation:

$\cos(\frac{\pi}{2})=0$ while $\sin(\frac{\pi}{2})=1$

So the equation becomes:

$-64=(2(0)+2i(1))^n$

$-64=(0+2i)^n$

$-64=(2i)^n$

We know that $2^6=64$. So let's see what $n=6$ gives us:

$(2i)^6=64i^6=64i^4i^2=64(1)(-1)=-64$.

$-64$  is the result we wanted.

$n=6$ is therefore a solution.