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(1 - i)^7 use de moivre's theorem

 Oct 10, 2014

Best Answer 

 #1
avatar+33661 
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cos(pi/4) = 1/√2   sin(pi/4) = 1/√2

 

1 - i = √2(cos(pi/4) - i*sin(pi/4))

 

(1 - i)7 = (√2)7(cos(pi/4) - i*sin(pi/4))7

 

De Moivre's theorem says  (cos(x) - i*sin(x))n = cos(nx) - i*sin(nx) so:

(1 - i)7 = (√2)7(cos(pi/4) - i*sin(pi/4))7  = 8√2*(cos(7pi/4) - i*sin(7pi/4)) = 8√2*(1/√2 - i*(-1/√2))

 

or (1 - i)7 =  8(1 + i)

 

.

 Oct 10, 2014
 #1
avatar+33661 
+5
Best Answer

cos(pi/4) = 1/√2   sin(pi/4) = 1/√2

 

1 - i = √2(cos(pi/4) - i*sin(pi/4))

 

(1 - i)7 = (√2)7(cos(pi/4) - i*sin(pi/4))7

 

De Moivre's theorem says  (cos(x) - i*sin(x))n = cos(nx) - i*sin(nx) so:

(1 - i)7 = (√2)7(cos(pi/4) - i*sin(pi/4))7  = 8√2*(cos(7pi/4) - i*sin(7pi/4)) = 8√2*(1/√2 - i*(-1/√2))

 

or (1 - i)7 =  8(1 + i)

 

.

Alan Oct 10, 2014
 #2
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thx a lot :)

 Oct 11, 2014

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