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

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### 2 Answers

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**

We suppose that z' is the conjugate of the complex number z.

z = a + bi

z' =a - bi

We know that z + z' = a+bi + a-bi

We'll eliminate like terms and we'll get:

z + z' = 2a (1)

We'll calculate the product:

z*z' = (a+bi)(a-bi) = a^2 - (bi)^2

z*z' = a^2 + b^2 (2)

We'll factorize by z*z' to the right side:

z+z'= z*z'(z + i)

We'll substitute the sum and the product by (1) and (2).

2a = (a^2 + b^2)*(z + i)

We'll divide by (a^2 + b^2):

2a / (a^2 + b^2) = z + i

We'll substitute z by a + bi and we'll calculate z+i.

z+i = a + bi + i

We'll factorize by i:

z + i = a + i(b+1)

The equation will become:

2a / (a^2 + b^2) = a + i(b+1)

Comparing, we'll get:

The real parts from both sides:

2a / (a^2 + b^2) = a

and

The imaginary parts:

b+1 = 0

We'll subtract 1:

b = -1

We'll determine a:

2a / (a^2 + b^2) = a

We'll divide by a both sides:

2 / (a^2 + b^2) = 1

a^2 + b^2 = 2

Since b = -1

a^2 + 1 = 2

We'll subtract 1:

a^2 = 1

a1 = 1

a2 = -1

**The solutions of the equation are: {1 - i ; -1 - i }**.