Question

(a) Show that, for any complex number z, zz = |z|
2
, z + z = 2Re(z) and Re(z) ≤ |z|. Hence
show that
i. |z1 + z2|
2 = |z1|
2 + |z2|
2 + 2Re(z1z2),
ii. |z1 + z2| ≤ |z1| + |z2|,
where Re(z) is the real part of z and z the conjugate of z.

Answer 1

a) Let $z=x+iy$ , then:

$z\cdot \overline {z}=(x+iy)(x-iy)=x^2+y^2=|z|^2$

$z+\overline z=x+iy+x-iy=2x=2Re(z)$

$Re(z)=x=\sqrt{x^2}\leq|z|=\sqrt{x^2+y^2}$

i) Let $z_1=x_1+iy_1,z_2=x_2+iy_2$ , then:

$|z_1+z_2|^2=(x_1+x_2)^2+(y_1+y_2)^2=(x_1^2+y_1^2)+(x_2^2+y_2^2)+2(x_1x_2+y_1y_2)=$

$=|z_1|^2+|z_2|^2+2Re(z_1z_2)$

ii)

$|z_1+z_2|=\sqrt{(x_1^2+y_1^2)+(x_2^2+y_2^2)+2(x_1x_2+y_1y_2)}$

$|z_1|+|z_2|=\sqrt{x_1^2+y_1^2}+\sqrt{x_2^2+y_2^2}$

$|z_1+z_2|^2=(x_1+x_2)^2+(y_1+y_2)^2=(x_1^2+y_1^2)+(x_2^2+y_2^2)+2(x_1x_2+y_1y_2)$

$(|z_1|+|z_2|)^2=(x_1^2+y_1^2)+(x_2^2+y_2^2)+2\sqrt{x_1^2x_2^2+y_1^2y_2^2+x_1^2y_2^2+x_2^2y_1^2}$

Since

$x_1x_2+y_1y_2\leq\sqrt{x_1^2x_2^2+y_1^2y_2^2+x_1^2y_2^2+x_2^2y_1^2}$

then

$|z1 + z2| ≤ |z1| + |z2|$