Solve-> z^2 + 2(1+j)z + 2 = 0, (complex number), how to solve?

Question

Solve-> z^2 + 2(1+j)z + 2 = 0,
in form a+jb

ans:-0.36+j0.55, -1.64-j2.55

show me the wroking plz..
tq and regards.

Answer 1

Others have tried but most of them have not been able to complete

comparing with ax^2+bx+c = 0

b^2- 4ac = 4(1+j)^2 - 8
= 4(1-1+2j) -8
= 8(j-1)

fortunately a and c are real so good

now we need to find square root of b^2-4ac
= 8(-1+j)

to find square root say re^(it) = -1 + j

r = sqrt(1+1) = sqrt(2)

cos t = -1/sqrt(2)

sin t = 1/ sqrt(2)

t = 3pi/4
b^2-4ac = 8 sqrt(2) e^(3pi/4)
sqrt(b^2-4ac) = sqrt(8sqrt(2)) e^(3pi/8)

= 2sqrt(2sqrt(2))(cos 3pi/8 + j sin 3pi/8)

now z = (-(1+j) +/-2sqrt(2sqrt(2))(cos 3pi/8 + j sin 3pi/8))/2
now I have solved it keeping the real and imaginary parts separatedly and a couple of steps from here shall take you

z1= (-1-2 sqrt(2sqrt(2))cos (3pi/8)/2 +j(-1 - 2 sqrt(2sqrt(2)) sin 3pi/8)
z2= (-1+2 sqrt(2sqrt(2))cos (3pi/8)/2 +j(-1 + 2 sqrt(2sqrt(2)) sin 3pi/8)

sf2h,,,, above started well and intention is very good but has put a complex number under square root and problem is not solved till square root is evaluated

Answer 2

z=x+jy

z^2 = (x^2 - y^2) + 2jxy


(1+j)z = (x-y) + j(x+y)

therefore, the given equation becomes

(x^2 - y^2)+(2*(x-y))+2 + j(2xy+2(x+y)) = 0


equating the real and imaginary parts

(x^2 - y^2)+(2*(x-y))+2 = 0

and

(2xy+2(x+y)) = 0


(2xy+2(x+y)) = 0 => xy+x+y=0

x(1+y) = -y

x= -y/(1+y)

Substitute this in

(x^2 - y^2)+(2*(x-y))+2 = 0

U'll get the answer

Answer 3

Use a combination of polar and cartesian representations of complex numbers and the quadratic equation.

Answer 4

z^2 + 2(1+j)z + 2 = 0

z^2 + (2+2j)z + 2 = 0

Quadratic formula:

z = [-(2+2j) +/- sqrt( (2+2j)^2 - 4*1*2 ) ] / 2*1

z = [ -2-2j +/- sqrt( (2+2j)^2 - 8 ) ] / 2

z = [ -2-2j +/- sqrt( (4 + 8j - 4) - 8 ) ] / 2

z = (-1 - j) +/- sqrt(2j - 2)

z = -1 - j + sqrt(2j - 2)
z = -1 - j - sqrt(2j - 2)

Those should be the same as the answers you've provided.

Answer 5

Exactly what he (the guy before me) said.

Answer 6

er....like the guys above me said

Answer 7

Or, do your own homework and learn from it.