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Thursday, June 15, 2023

Challenge:Complex reducible to quadratic equation: Solve z4+2(a2b2)z2+(a2+b2)2=0 where a,bR

Solve z4+2(a2b2)z2+(a2+b2)2=0 where a,b are real numbers.

Hint 1 : Put w=z2 and find w first.  Then for each w, solve z2=w.

Hint 2: see other posts on this label.




























Answer: z=±(a+ib),±(aib)

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