+50 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.
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.
+373 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
+26367 Find all complex numbers z such that z^4=-4
here is another answer: https://web2.0calc.com/questions/please-help_20319#r2
