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Find all complex numbers \(z\) such that \(z^4 = -4.\)

Note: All solutions should be expressed in the form \(a+bi,\) where \(a\) and \(b\) are real numbers.

 Aug 10, 2021
 #1
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We can write -4 in exponential notation as 4e^(pi*i), so the equation is z^4 = 4e^(pi*i).

 

By Hamilton's Theorem, the solutions are z = 4^{1/4}*e^(pi*i/4), 4^{1/4}*e^(pi*i/4 + pi/4), 4^{1/4}*e^(pi*i/4 + 2*pi/4), and 4^{1/4}*e^(pi*i/4 + 3*pi/4).  Since 4^{1/4} = sqrt(2) and e^(pi*i/4) = (1 + i)/sqrt(2), the first solution is 1 + i.  Then the other roots work out as

 

4^{1/4}*e^(pi*i/4 + pi/4) = 1 - i,

4^{1/4}*e^(pi*i/4 + 2*pi/4) = -1 - i, and

4^{1/4}*e^(pi*i/4 + 3*pi/4) = -1 + i.

 Aug 10, 2021
 #2
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i don't quite get it, is there another way?

 Aug 11, 2021
 #3
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Hi justinwh333,

 

Your question is already answered here: https://web2.0calc.com/questions/please-help_20319#r2 

 Aug 11, 2021

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