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

 Dec 16, 2019
 #1
avatar+19741 
0

z^4 = -4

z^2 = +- sqrt(-4) = +-2i

z = +- sqrt(+-2i)

=  -1+i  ,   -1-i    ,  1-i    ,  1+i 

 Dec 16, 2019
 #2
avatar+43 
-1

For styling purposes:

\(\pm(1+i), \pm(1-i)\)

whoisjoe  Dec 16, 2019
 #5
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+1

Question: is there a way to prove that those are ALL of the numbers? Thanks

Guest Dec 17, 2019
 #6
avatar+106885 
+1

I personally do not know how to prove it but with all roots there is always 'that' number of roots,

so square root has 2 values

cube roots have 3 values

4th roots have 4 values  etc

 

I am sure if you google it you will find a proof.

Melody  Dec 17, 2019
 #7
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+1

Got it. Thanks Melody!

Guest Dec 18, 2019
 #3
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+1

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.

 Dec 17, 2019
 #4
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+1

that... is a more complicated way. BUT AWESOME! 

Guest Dec 17, 2019

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