(1+i)z=64i/3

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Solution for (1+i)z=64i/3 equation: 


x in (-oo:+oo)

z*(i+1) = (64*i)/3 // - (64*i)/3

z*(i+1)-((64*i)/3) = 0

z*(i+1)+(-64/3)*i = 0

z*(i+1)+(-64*i)/3 = 0

(3*z*(i+1))/3+(-64*i)/3 = 0

3*z*(i+1)-64*i = 0

3*i*z-64*i+3*z = 0

(3*i*z-64*i+3*z)/3 = 0

(3*i*z-64*i+3*z)/3 = 0 // * 3

3*i*z-64*i+3*z = 0

x belongs to the empty set

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