How do you get the 4 roots of z^4 +4=0?

Question

I know that if z^4+4=0

Then z^2 = +/- 2i

Then I can use z=x+iy but I just can't get the correct answer, I have the answers

They are

1+i
1-i
-1+i
-1-i

can any one let me know how to use z=x+iy to solve?

Answer 1

z² = ±2i

(x² - y²) + 2xyi = ±2i

Equating the real parts:

x² - y² = 0, so x² = y²

Equating the imaginary parts:

2xyi = ±2i

so xy = ±1.

From the first equation x = ±y
From the second equation xy = ±1

If x has to be the same size as y then the only pairs of x and y that multiply to give 1 or -1 are:

(1, 1) ; (-1, -1); (-1, 1); (1, -1)

Therefore, the solutions are 1+i, -1-i, -1+i and 1-i.

Answer 2

we have got z^2 = +/-2 i

there are 2 ways

z^2 = + 2i
let z = (x+iy) where x and y are real
z^2 = x^2-y^2 + 2ixy
equating real part and imaginary parts

x^2-y^2 = 0 so x = +/-y
2xy = 2 gives xy = 1
x= y gives x^2 = 1 so x = +/- 1
this gives 2 solutions 1+ i and -1 - i

x = - y gives x^2 = -1 which is not possible(as x is real)

similarly you can solve z^2 = -2i

But there is a better formula if you like
using e^it = cos t + i sin t

z^4 = -4
let - 4 = r cos t + i r sin t
r = 4

cos t = -1 and sint = 0 so t = pi

z^4 = 4 cos pi + 4 i sin pi = 4e^(ipi)

z = 4^(1/4) e^i(2npi+pi)/4 = 2^(1/2)(cos (2n+1)/4pi + i sin (2n+1)/4
putitng n = 0 to 3 you get z
n= 1gives
= 2^(1/2)(cos pi/4 + i sin pi/4)
= 1 + i
similarly for other n
n =5 shall give same as n= 1



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Answer 3

Here's the easy way. Just think of complex numbers in polar form. Then the equation is written as

z^4 = -4 = 4 expi(pi)

Taking the fourth root is the same as taking the 1/4 power of the modulus, and 1/4 of the angle. For roots, you additionally space the four roots evenly around the circle. So the principal root is 4^(1/4) expi(pi/4) = sqrt 2 * expi(pi/4). The other three roots have the same modulus, sqrt 2, with angles 3pi/4, 5pi/4, and 7pi/4.

So, putting the numbers back into rectangular coordinates, the four roots are

1 + i, -1 + i, -1 - i, and 1 - i.

Answer 4

z^2 = (x + iy)^2 = x^2 + 2xyi - y^2

In this case:
x^2 + 2xyi - y^2 = 2i

So case 1:
x^2 - y^2 =0 ====> lxl = lyl
2xy = 2 ==> xy = 1

x^2 - (1/x)^2 = 0
x^4 = 1

x = 1 or
x = -1

In which case:
y=1 or
y = -1

Applied, the 4 roots are:
1+i
1-i
-1+i
-1-i

Answer 5

Working in trigonometric form
z^4=4 so z=4^1/4 < pi/4 +kpi/2 with k(0,1,2,3)
In binomial form it would be
z=4^1/4(cos(pi/4+kpi/2)+i sin(pi/4+kpi/2)