## Example 1.1.9 (d): Algebra with complex numbers |

Solve the equations |

We represent *z = x + iy* and expand the first equation:

z^{2}= (x + iy)^{2}= x^{2}- y^{2}+ i 2xy = i

Now we equate like terms (real and imaginary parts of the equation) to arrive at a system of two equations with two unknowns:

(1) x^{2}- y^{2}= 0

(2) 2xy = 1

The first equation implies that *x = y* or *x = -y*.
Using this in the second equation gives
*x = * or
*x = -*.
Thus, the two roots of *i* are:

z_{1}= + i

z_{2}= - - i

We proceed similar for the second equation:

z^{2}= (x + iy)^{2}= x^{2}- y^{2}+ i 2xy = 1 + 2i

Again we equate like terms to arrive at a system of two equations with two unknowns:

(1) x^{2}- y^{2}= 1

(2) 2xy = 2

Our system of equations is a little harder than before, but it's not too
bad. Solving the second equation for *y* (we can exclude
*x=0* as a possible solution) gives *y = 1/x*.
We can substitute that into the first equation to get:

x^{2}- 1/x^{2}= 1

We multiply through with *x ^{2}* and bring everything on
one side:

x^{4}- 1 - x^{2}= 0

You should have no trouble solving this equation, using the quadratic
formula and a substitution like *u = x ^{2}*. You

z_{1}= + i

z_{2}= - - i