# complex number question

• Jul 27th 2009, 06:01 PM
stud_02
complex number question
$\displaystyle Z = \frac{8(1 + i)}{\sqrt{2}}$

write Z as r(cosΘ +i sinΘ)[/tex]

using that show the three values for $\displaystyle Z^{2/3}$ are

$\displaystyle -4i , 2(\sqrt{3} + i ) ,2(-3\sqrt{3} + i )$

----
i can do the first part

$\displaystyle z=8(cos(\pi/4) + i sin(\pi/4))$

and since $\displaystyle z^n =8^n (cos( n* \pi/4) + i sin(n*\pi/4))$

$\displaystyle z^{2/3} = 4( cos\pi/6 + isin\pi/6)=2(\sqrt{3} + i)$

i cant get other 2 numbers
• Jul 27th 2009, 07:02 PM
HallsofIvy
Quote:

Originally Posted by stud_02
$\displaystyle Z = \frac{8(1 + i)}{\sqrt{2}}$

write Z as r(cosΘ +i sinΘ)[/tex]

using that show the three values for $\displaystyle Z^{2/3}$ are

$\displaystyle -4i , 2(\sqrt{3} + i ) ,2(-3\sqrt{3} + i )$

----
i can do the first part

$\displaystyle z=8(cos(\pi/4) + i sin(\pi/4))$

and since $\displaystyle z^n =8^n (cos( n* \pi/4) + i sin(n*\pi/4))$

$\displaystyle z^{2/3} = 4( cos\pi/6 + isin\pi/6)=2(\sqrt{3} + i)$

i cant get other 2 numbers

Since sine and cosine are periodic with period $\displaystyle 2\pi$, z is also equal to $\displaystyle z= 8(cos(\pi/4+ 2\pi)+ i sin(\pi/4+ 2\pi))$ and $\displaystyle z= 8(cos(\pi/4+ 4\pi)+ i sin(\pi/4+ 4\pi))$.

Apply that same formula to this.