Victoriaflower
Victoriaflower Victoriaflower
  • 22-11-2020
  • Mathematics
contestada

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

Respuesta :

freckledspots freckledspots
  • 22-11-2020

Answer:

The real solution is [tex]n=6[/tex].

Step-by-step explanation:

[tex]\cos(\frac{\pi}{2})=0[/tex] while [tex]\sin(\frac{\pi}{2})=1[/tex]

So the equation becomes:

[tex]-64=(2(0)+2i(1))^n[/tex]

[tex]-64=(0+2i)^n[/tex]

[tex]-64=(2i)^n[/tex]

We know that [tex]2^6=64[/tex]. So let's see what [tex]n=6[/tex] gives us:

[tex](2i)^6=64i^6=64i^4i^2=64(1)(-1)=-64[/tex].

[tex]-64[/tex]  is the result we wanted.

[tex]n=6[/tex] is therefore a solution.

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