Answer to Question #124528 in Complex Analysis for Amoah Henry

Question

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. [26 marks]
(b) If z1 = 1 + 2i, find the set of values of z2 for which
(i) |z1 + z2| = |z1| + |z2| (ii) |z1 + z2| = |z1| − |z2|.

Answer 1

As $|z|=\sqrt{x^2+y^2}$ where $z=x+iy,\bar{z}=x-iy,$ then $z\sdot\bar{z}=x^2+y^2=|z|^2$ and

$z+\bar{z}=2x=2Re(z)$ where $x=Re(z)$ , then $Re(z)\le |z|$

i. $|z_1+z_2|=(z_1+z_2)(\bar{z_1}+\bar{z_2})=z_1\bar{z_1}+z_1\bar{z_2}+z_2\bar{z_1}+z_2\bar{z_2}=$

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

ii. Since the sum of two complex numbers can be represented as the third side of a triangle then

$|z_1+z_2|\le|z_1|+|z_2|$

b) Using the triangle rule

i. If $|z_1+z_2|=|z_1|+|z_2|$ then $z_2=k(1+2i)$ where $k>0, k\in R$

ii. If $|z_1+z_2|=|z_1|+|z_2|$ then $z_2=-k(1+2i)$ where $k>0,k\in R$