complex number question

**stud_02** · July 27th 2009, 06:01 PM

$Z = \frac{8(1 + i)}{\sqrt{2}}$

write Z as r(cosΘ +i sinΘ)[/tex]

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

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

----
i can do the first part

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

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

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

i cant get other 2 numbers

**HallsofIvy** · July 27th 2009, 07:02 PM

Originally Posted by stud_02

$Z = \frac{8(1 + i)}{\sqrt{2}}$

write Z as r(cosΘ +i sinΘ)[/tex]

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

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

----
i can do the first part

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

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

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

i cant get other 2 numbers

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

Apply that same formula to this.

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