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

 Jan 13, 2020
 #1
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+2

You can search for this post to find the derivation as it was posted recently

 

-1 -i

-1+i

1-i

1+i

 Jan 13, 2020
 #2
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thx!

Guest Jan 14, 2020
 #3
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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.

 Jan 16, 2020
 #4
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Why can't you just take the square root, get +/- 2i and then take the square root again?

 Jan 17, 2020

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