So far I have only done:
z^4-24z^2+169=0
z^4-24z^2= -169
z^4-24z^2= 169 cis Pi
2007-08-07 18:31:55 · 4 answers · asked by cherrichoc 2 in Science & Mathematics ➔ Mathematics
what does iarctan mean??
2007-08-07 18:44:36 · update #1
Holy crap man what is that. Trig
2007-08-07 18:34:12 · answer #1 · answered by Chad 2 · 0⤊ 2⤋
Find all the roots of
z^4 - 24z² + 169 = 0
z^4 - 24z² = -169
(z^4 - 24z² + 144) = -169 + 144
(z² - 12)² = -25
z² - 12 = ±√(-25) = ± 5i
z² = 12 ± 5i
z = ±√(12 ± 5i)
There are four roots.
2007-08-07 18:50:49 · answer #2 · answered by Northstar 7 · 2⤊ 0⤋
(z^2-12)^2 = -25
z^2 =12±5i = 13e^(iarctan(±5/12)+2npi*i)
z = √13e^[(iarctan(±5/12)
+2npi*i)/2], n = 0, 1
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There are four roots.
2007-08-07 18:41:15 · answer #3 · answered by sahsjing 7 · 1⤊ 0⤋
homework is stupid
2007-08-07 18:34:16 · answer #4 · answered by Anonymous · 0⤊ 2⤋