Solve the equation z+z'=z'*z^2+i*z*z' where z is a complex number.

Expert Answers

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As z is a complex number z = x+ iy

z' = x - iy

z + z' = x + iy + x - iy = 2x

z'*z^2 + i*z*z' = (x - iy)*(x +iy)^2 + i*(x - iy)(x + iy)

=> (x^2 + y^2)(x + iy) + i(x^2 + y^2)

=> x^3 + x^2yi + y^2x + y^3i + x^2i + y^2i

=> x^3 + xy^2 + x^2yi + y^3i + x^2i + y^2i

This is equal to 2x

=> x^3 + xy^2 + x^2yi + y^3i + x^2i + y^2i = 2x

=> x^3 + xy^2 = 2x

=> x^2 + y^2 = 2

x^2y + y^3 + x^2 + y^2 = 0

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

=> 2y + 2 = 0

=> y = -2/2

=> y = -1

x^2 = 2 - y^2 = 1

x = 1 , -1

Therefore z = 1 - i and -1 - i

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