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Find all complex numbers $z$ such that $z^4 = -4.$

 May 11, 2020
 #1
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Let's make the solution super-formal!  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.

 May 12, 2020
 #2
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-1

I tried plugging in your answers into the equation, but they didn't quite work... I did find out that -1-i, and 1+i do work though.

trumpstinks  May 12, 2020

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