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

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

Answer 1

Convert the 1+j to exp(j*ang) and use quadratic formula. That's because the sqrt is very easy with the exponential form.

In other words, the trick here is to use a combination of polar and cartesian representations of complex numbers and the quadratic equation.

Answer 2

z=x+jy z^2 = (x^2 - y^2) + 2jxy (a million+j)z = (x-y) + j(x+y) subsequently, the given equation will change into (x^2 - y^2)+(2*(x-y))+2 + j(2xy+2(x+y)) = 0 equating the authentic 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(a million+y) = -y x= -y/(a million+y) change this in (x^2 - y^2)+(2*(x-y))+2 = 0 U'll get the answer