Complex Analysis

“ca: chapter_01” — 2012/3/27 — 20:27 — page 19 — #19

Complex Numbers

|

19

x

z

r

r

y

O

1

θ

–θ

Fig. 1.19

Example 1.9: (a) Prove that arg(z) − arg(−z) =±π.

(b) Show that arg z + arg

z = 2nπ , where n is an integer.

Solution: (a) By the property of argument, we have

arg



z

1

z

2



= arg(z

1

) − arg(z

2

)

and arg(−c) =±π , where cis positive real number.

Therefore, we get arg(z) −arg(−z) = arg



z

−z



= arg(−1) =±π.

(b) Let z = x + iy, then

z = x − iy

Now, we know that

arg z + arg

z = arg zz = arg

[

(x + iy)(x − iy)

]

= arg(x

2

+ y

2

)

= arg a, wherea = x

2

+ y

2

Since a is positive real number so in polar form a = rcosθ ,0 = rsinθ

⇒ r = a and cos θ = 1, sin θ = 0

∴ The general value of θ is 2nπ,