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The number \(16\) has four fourth roots. In other words, there are four complex numbers that can be entered in the square in the equation below: \(\square^4=16.\) Find them.

 

I know that 2 is one, but I don't know how to find the others. Could anyone please provide an explanation with their answer? Thank you!

 Jul 1, 2020
 #1
avatar+711 
+2

hi otterstar!

 

so we know that \((-2)^4=16\) since -2*-2*-2*-2=16

 

since there are 4 roots, and we've already covered all the real ones, we have to look for complex numbers now.

 

we also know that \(i^4=1\), so that means that \((2i)^4=16\) since 2^4 already equals 16, and multiplying it by 1 will equal 16 too.

 

lastly, we know that \((-2i)^4=16\)  since we already know that -2^4 equals 16 (as we found earlier) and multiplying it by i^4=1 will equal 16 too. 

 

so the solutions are \(\boxed{2,-2,2i,-2i}\)

 

 

 

i hope this helped you! :)

 Jul 1, 2020
 #2
avatar+132 
+1

Thank you so much!

Otterstar  Jul 1, 2020
 #3
avatar+711 
+1

no problem! :)

lokiisnotdead  Jul 1, 2020

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