SOLUTION: By expressing (1+j) in polar form and using De Moivre's Theorem, show that (1+j)^24 = 2^12. Obtain (1-j)^24. Show that (2 sqrt3- 2j)^12 =4^12. Obtain (2 sqrt3+2j)^12 Thanks

Question 402776: By expressing (1+j) in polar form and using De Moivre's Theorem, show that (1+j)^24 = 2^12. Obtain (1-j)^24.
Show that (2 sqrt3- 2j)^12 =4^12. Obtain (2 sqrt3+2j)^12
Thanks
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
By expressing (1+j) in polar form and using De Moivre's Theorem, show that (1+j)^24 = 2^12.
1+j --> sqrt(2)cis(45�)
(1+j)^24 = (sqrt(2))^24 cis(45*24) = 2^12 cis (1080)
= 2^12
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Obtain (1-j)^24.
I don't know what that means. If it means calculate, do it like the one above.
===========================
Show that (2 sqrt3- 2j)^12 =4^12.
Similar to the one above.
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Obtain (2 sqrt3+2j)^12
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email via the thank you note and I'll check your work.

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