Solve the equation : 2│z| ^2 + z^ 2 – 5 + i√3 = 0.
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we know
|z|=root (x^2+y^2)
where x is real number and y is imaginary number
now
|z|^2+z^2-5+iroot3=0
=> x^2+y^2+(x+iy)(x+iy)-5+iroot3=0
=> x^2+y^2+x^2-y^2+2ixy-5+iroot3=0
=>(2x^2-5)+i(2xy+root3)=0
now
2x^2-5=0
x=+_root (5/2)
and also 2xy+root3=0 put x value and find y
y=_+root (3/10)
|z|=root (x^2+y^2)
where x is real number and y is imaginary number
now
|z|^2+z^2-5+iroot3=0
=> x^2+y^2+(x+iy)(x+iy)-5+iroot3=0
=> x^2+y^2+x^2-y^2+2ixy-5+iroot3=0
=>(2x^2-5)+i(2xy+root3)=0
now
2x^2-5=0
x=+_root (5/2)
and also 2xy+root3=0 put x value and find y
y=_+root (3/10)
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