Show that |z^2| = a^2 + b^2 algebraically?

Question

please Help!!!!!!!!!!!!!!!!!

 

z is a complex number

Answer 1

z is a compiex number

z = a + i.b

Where i² = - 1 , The real number a is called the real part of z, and the real number b is called the imaginary part of z.

i.e.

a = Re(z) , b = Im(z).

 

z = r.cos theta + i.r.sin theta,

where r² = x² + y², cos theta = x/r, and sin theta = y/r. The real number r is called the absolute value of z, and the real number theta is called the argument of z.

i.e.

r = | z |, theta = arg(z).

Now it can be show that

| z² | = a² + b²

    z = a + i.b

We know that  r = | z |   and  r² = x² + y²

So  r² = | z² | = a² + b²

There fore | z² | = a² + b²