Math 411: Honours Complex Variables - University of Alberta

Proof. (i): If z = x+iy, then ¯z = x−iy, so that 2x = z+¯z; this yields the claim for
Rez. The assertion for Imz is proven similarly.
(ii) is obvious.
(iii): Let z = x+iy and w = u+iv, so that
and thus
zw = (xu−yv)+i(xv +yu)
zw = (xu−yv)−i(xv +yu).
On the other hand, we have ¯z = x−iy and ¯w = u−iv, which yields
as claimed.
(iv): By (iii), we have
which yields the claim.
¯z¯w = (xu−(−y)(−v))+i(x(−v)+(−y)u)
= (xu−yv)−i(xv +yu)
= zw,
z −1 ¯z = z −1 z = ¯1 = 1,
For any z = x+iy ∈ C, we note that z¯z = x 2 +y 2 ≥ 0. This provides us with a
natural generalization of the absolute value function to C.
Definition. For z ∈ C, set |z|:= √ z¯z.
Proposition 1.3. |·| is the Euclidean norm on R 2 . In particular, the following hold:
(i) |z|≥ 0 with |z|= 0 if and only if z = 0;
(ii) |z +w|≤ |z|+|w| for z,w ∈ C.
Moreover, we have |zw|= |z||w| for z,w ∈ C and z −1 = ¯z
|z| 2 for z ∈ C\{0}.
Proof. Noting that |x+iy|= � x 2 +y 2 , we see that |·| is the Euclidean norm, which
entails (i) and (ii). Letting z,w ∈ C, we see that
|zw| 2 = zwzw = (z¯z)(w¯w) = |z| 2 |w| 2 .
Also, since |z| 2 = z¯z, we have 1 = z ¯z
|z| 2 for z �= 0 and thus z −1 = ¯z
|z| 2.
There is a remarkable similarity between the complex multiplication rule
(x,y)·(u,v) = (xu−yv,xv+yu)
7