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Find all (real or nonreal)  z satisfying

\((z - 3)^4 + (z - 5)^4 = -8\)

 Jul 22, 2024
 #1
avatar+1908 
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First, let's expand out everything. I'm not going to show the steps, but we eventually get

\(2z^4-32z^3+204z^2-608z+706=-8\\2z^4-32z^3+204z^2-608z+714=0\)

 

Now, since every number is divisible by 2, we get factor out 2 to get

\({2\left(z^{4}-16z^{3}+102z^{2}-304z+357\right)}=0\)

 

Now, note that \(z^{4}-16z^{3}+102z^{2}-304z+357=(z^{2}-8z+17)(z^{2}-8z+21)\)

 

This means that we have the equation

\(2\left(z^{2}-8z+17\right)\left(z^{2}-8z+21\right)=0\)

 

Since the equation is equal to 0, we get two equations. We have

\(z^{2}-8z+17=0,\qquad z^{2}-8z+21=0\)

 

Using the quadratic equation on both formulas, we find 4 values of z. 

We have

\(z=4+i\\ z=4-i\\ z=4+\sqrt{5}\cdot (i)\\ z=4+\sqrt{5}\cdot (-i)\)

 

This was a very basic explenation of the problem. 

I can go in dephth more if you want to. 

 

Thanks! :)

 Jul 22, 2024
edited by NotThatSmart  Jul 22, 2024
 #2
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Thank you! I just really want the methods.

MeldHunter  Jul 23, 2024

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