Hi Gibsonj338
Find all cubic roots of
z=−1+i distance of z to the origin =√12+12=√2cubic toots so there are 3 of them so they are 2π3radiansapart.z=√2(cis3π4)3√z=6√2(cis(3π4÷3)),6√2(cis((3π4+2π)÷3)),6√2(cis((3π4+4π)÷3))3√z=6√2(cis(3π4÷3)),6√2(cis(11π4÷3)),6√2(cis(19π4÷3))3√z=6√2(cis(3π12)),6√2(cis(11π12)),6√2(cis(19π12))3√z=6√2(cis(π4)),6√2(cis(11π12)),6√2(cis(19π12))or
3√z=1+i3√2,6√2(cos(11π12)+isin(11π12)),6√2(cos(19π12)+isin(19π12))3√z=1+i3√2,6√2(−cos(π12)+isin(π12)),6√2(cos(5π12)−isin(5π12))
z=(−1+i)(1/3)
r=√((−1)2+12)
r=√(1+12)
r=√(1+1)
r=√(2)
tan(Θ)=1/−1
tan(Θ)=−1
Θ=tan−1(−1)
Θ=−π/4
√(2)e(−π/4i)(1/3)
√(2)(1/3)e(−π/12i)
√(2)(1/3)∗(cos(−π/12)+isin(−π/12))
√(2)(1/3)∗(0.965925826289+isin(−π/12))
√(2)(1/3)∗(0.965925826289+i∗−0.258819045103)
1.414213562373095)(1/3)∗(0.965925826289+i∗−0.258819045103)
1.122462048309373∗(0.965925826289+i∗−0.258819045103)
1.084215081491274550528506797+i∗−0.290514555507789375384650419
1.084215081491274550528506797+−0.290514555507789375384650419i
Answer 1: 1.084215081491274550528506797−0.290514555507789375384650419i
√(2)e(7π/4i)(1/3)
√(2)(1/3)e(7π/12i)
√(2)(1/3)∗(cos(7π/12)+isin(7π/12))
√(2)(1/3)∗(−0.258819045103+isin(7π/12))
√(2)(1/3)∗(−0.258819045103+i∗0.965925826289)
1.414213562373095)(1/3)∗(−0.258819045103+i∗0.965925826289)
1.122462048309373∗(−0.258819045103+i∗0.965925826289)
−0.290514555507789375384650419+i∗1.084215081491274550528506797
Answer 2: −0.290514555507789375384650419+1.084215081491274550528506797i
√(2)e(15π/4i)(1/3)
√(2)(1/3)e(15π/12i)
√(2)(1/3)∗(cos(15π/12)+isin(15π/12))
√(2)(1/3)∗(−0.707106781187+isin(15π/12))
√(2)(1/3)∗(−0.707106781187+i∗−0.707106781187)
1.414213562373095)(1/3)∗(−0.707106781187+i∗−0.707106781187)
1.122462048309373∗(−0.793700525984607637192165751+i∗−0.793700525984607637192165751)
−0.793700525984607637192165751+i∗−0.793700525984607637192165751
−0.793700525984607637192165751+−0.793700525984607637192165751i
Answer 3: −0.793700525984607637192165751−0.793700525984607637192165751i
Hi Gibsonj338
Find all cubic roots of
z=−1+i distance of z to the origin =√12+12=√2cubic toots so there are 3 of them so they are 2π3radiansapart.z=√2(cis3π4)3√z=6√2(cis(3π4÷3)),6√2(cis((3π4+2π)÷3)),6√2(cis((3π4+4π)÷3))3√z=6√2(cis(3π4÷3)),6√2(cis(11π4÷3)),6√2(cis(19π4÷3))3√z=6√2(cis(3π12)),6√2(cis(11π12)),6√2(cis(19π12))3√z=6√2(cis(π4)),6√2(cis(11π12)),6√2(cis(19π12))or
3√z=1+i3√2,6√2(cos(11π12)+isin(11π12)),6√2(cos(19π12)+isin(19π12))3√z=1+i3√2,6√2(−cos(π12)+isin(π12)),6√2(cos(5π12)−isin(5π12))