a⁴=-64 Let a=x+yi {x,y∈R} (x+yi)⁴=-64 x⁴+4x³yi-6x²y²-4xy³i+y⁴=-64+0i Comparing the real and imaginary parts of both sides of the equation x⁴-6x²y²+y⁴=-64 ① and 4x³y-4xy³=0 4xy(x²-y²)=0 4xy(x-y)(x+y)=0 => 4 cases case 1: x=0 (rejected: from ① y⁴=-64, but y∈R) case 2: y=0 (rejected: from ① x⁴=-64, but x∈R) case 3: x-y=0 => x=y => from ①: -4y⁴=-64 => y⁴=16 => y=±⁴√16=±2 (complex roots rejected, b/c b∈R) if y=2 ==> x=2 => a₁=2+2i if y=-2 ==> x=-2 => a₂=-2-2i case 4: x+y=0 => x=-y => from ①: -4y⁴=-64 => y⁴=16 => y=±⁴√16=±2 (complex roots rejected, b/c b∈R) if y=2 ==> x=-2 => a₃=-2+2i if y=-2 ==> x=2 => a₄=2-2i
x^4=(2^6)(-1)=(2^6)e^{i2jπ}e^{iπ+i2jπ}=(2^6)e^{iπ+i2jπ}, j any integer and i=√(-1) x_j=(2^6)^{1/4}[e^{iπ+i2jπ}]^{1/4}=2^{3/2}e^{iπ/4+ijπ/2}=2^{3/2}[cos(π/4+jπ/2)+i*sin(π/4+jπ/2)],j=0,1,2,3 x_0=2√2(√2/2+i*√2/2)=2+2i, x_1=-2+2i, x_2=-2-2i, and x_3=2-2i. That is, the four roots are equally spaced on a circle of radius 2√2 centered on the origin of the complex plane.
a⁴=-64
Let a=x+yi {x,y∈R}
(x+yi)⁴=-64
x⁴+4x³yi-6x²y²-4xy³i+y⁴=-64+0i
Comparing the real and imaginary parts of both sides of the equation
x⁴-6x²y²+y⁴=-64 ①
and
4x³y-4xy³=0
4xy(x²-y²)=0
4xy(x-y)(x+y)=0 => 4 cases
case 1: x=0 (rejected: from ① y⁴=-64, but y∈R)
case 2: y=0 (rejected: from ① x⁴=-64, but x∈R)
case 3: x-y=0 => x=y =>
from ①: -4y⁴=-64 => y⁴=16 => y=±⁴√16=±2 (complex roots rejected, b/c b∈R)
if y=2 ==> x=2 => a₁=2+2i
if y=-2 ==> x=-2 => a₂=-2-2i
case 4: x+y=0 => x=-y =>
from ①: -4y⁴=-64 => y⁴=16 => y=±⁴√16=±2 (complex roots rejected, b/c b∈R)
if y=2 ==> x=-2 => a₃=-2+2i
if y=-2 ==> x=2 => a₄=2-2i
x^4=(2^6)(-1)=(2^6)e^{i2jπ}e^{iπ+i2jπ}=(2^6)e^{iπ+i2jπ}, j any integer and i=√(-1)
x_j=(2^6)^{1/4}[e^{iπ+i2jπ}]^{1/4}=2^{3/2}e^{iπ/4+ijπ/2}=2^{3/2}[cos(π/4+jπ/2)+i*sin(π/4+jπ/2)],j=0,1,2,3
x_0=2√2(√2/2+i*√2/2)=2+2i, x_1=-2+2i, x_2=-2-2i, and x_3=2-2i.
That is, the four roots are equally spaced on a circle of radius 2√2 centered on the origin of the complex plane.
a⁴=-64
a⁴+64=0
(a²)²-(8i)²=0
(a²-8i)(a²+8i)=0
(a²-(2√2√i)²)(a²-(2√2i√i)²)=0 but √i=1/√2+i/√2 (principal root)
(a-2√2(1/√2+i/√2))(a+2√2(1/√2+i/√2))(a-2√2i(1/√2+i/√2))(a+2√2i(1/√2+i/√2))=0
(a-2-2i)(a+2+2i)(a+2-2i)(a-2+2i)=0
so
a₁=2+2i
a₂=-2-2i
a₃=-2+2i
a₄=2-2i