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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.

 May 9, 2021
 #1
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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 9, 2021
 #2
avatar+234 
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Hey there, Chocolatehere!

 

FYI, this question has already been answered here: https://web2.0calc.com/questions/find-all-complex-numbers-z-such-that-z-4-4-note-all

 

So that's that.

Now have some chocolate. 🍫

 

Hope this helped! :)

( ゚д゚)つ Bye

 May 10, 2021
 #3
avatar+25988 
+2

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

 

here is another answer:  https://web2.0calc.com/questions/please-help_20319#r2

 

laugh

 May 10, 2021
 #4
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0

Thank you everyone!!!!

 May 14, 2021

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