Lobosito Marino replied"I already solved for
0.366(-1+j) and 0.366(1-j). "
how?

write z^2=t, so you have
t^2+4jt-1=0
now, use quadratic formula
after you find t, use Moivre formula


"complete the square like
z^4 + 4jz^2 -4 = 1-4
then (z^2 + 2j)^2 =-3"
GOOD
"from that factored form i solved 0.366(1-j) and
0.366(-1+j)"
How?

you should get this :
z^2+2j= (-3)^(1/2) = +/-j*sqrt(3)
so
z^2=-2j+/-j*sqrt(3)= (-2+/-sqrt(3))j

so you have
z^2=(-2+sqrt(3))j and
z^2=(-2-sqrt(3))j

let's choose for example z^2=(-2+sqrt(3))j =(2-sqrt(3))(-j)
so
z= ((2-sqrt(3))(-j) )^(1/2)
=sqrt(2-sqrt(3))*(-j)^(1/2)

Now, (-j)^(1/2) can be found with Moivre formula

"when I apply De Moivre I got
0.366(1-j) and 0.366(-1+j). "
I see,
Let me continue to see what I get
w=(-j)^(1/2)= (cos(3*pi/2)+j*sin(3*pi/2))^(1/2) which is by Moivre,
w_1=cos(3*pi/4)+j*sin(3*pi/4)=
-sqrt(2)/2+j*sqrt(2)/2
w_2=cos((3pi/2)+2pi)/2)+
j*sin((3pi/2)+2pi)/2)=
=sqrt(2)/2+j*sqrt(2)/2

so z_1=sqrt(2-sqrt(3)) *sqrt(2)/2*(-1+j)
z_2=sqrt(4-2sqrt(3))/2(1-j)


Now, you have to solve for
z^2=(-2-sqrt(3))j = (2+sqrt) (-j)

so
z=sqrt(2+sqrt(3))*(-j)^(1/2)
(-j)^(1/2) was found before, so we get
z_3=sqrt(2+sqrt(3))*
sqrt(2)/2*(-1+j)=
=sqrt(4+2sqrt(3))/2(-1+j)
z_4=sqrt(4+2sqrt(3))/2(1-j)
With approximation,
z_3=1.366(-1+j) and z_4=1.366(1-j)

for ernairnp: -0.366(-1+j)= 0.366(1-j), so by taking the negatives you get the same pair of solutions

ernairnp repliedOther two solutions are:

-0.366(-1+j) and -0.366(1-j)

intc_escapee repliedz^4 + 4iz² - 1 = 0
let u = z²
u² + 4iu - 1 = 0
u = (-4i ± √((4i)² + 4))/ 2 = i(-2 ± √3)
z² = i(-2 ± √3)
z = ± √(i(-2 ± √3))

Answer: z = ± √(i(-2 ± √3)) .... ≅ (0.366025 - 0.366025i), (-0.366025 + 0.366025i), (1.36602 -1.36602i), (-1.36602 + 1.36602i) in decimal

ron m repliedIf you are looking for roots, dig under the plants, and then check with your teacher on help understanding your homework.

JB repliedz^4 + 4jz^2 = 1
z^4 + 4jz^2 + (2j)^2 = 1+(2j)^2, but (2j)^2 = -4, so
(z^2+2j)^2 = -3, hence
z^2+2j = ±j√3, so
z^2 = (-2 ± √3) j = (1 ± √3/2)(-2j).

By De Moivre √(-2j) = 1-j, which is easy to check.
Also, ±√(1 ± √3/2) = ±(±1/2+√3/2), which you can check by squaring. Hence the four roots are:

z = ±(±1/2+√3/2)(1-j).