MCS 351 ENGINEERING MATHEMATICS SOLUTION OF ...

7. Let |z + 1| = |z − 1| and z = a + bi. If same things occur,
|z + 1| = |z − 1| ⇒ |(a + 1) + bi| = |(a − 1) + bi| ⇒ √ (a + 1) 2 + b 2 = √ (a − 1) 2 + b 2
⇒ (a + 1) 2 + b 2 = (a − 1) 2 + b 2
⇒ a = 0 ⇒ z = bi, b ∈ R
is found. Then we could picture it as following:
8. We now determine the region of |Argz| ≤ 1 4π. For this reason, z = x + iy is taken.
|Argz| ≤ 1 4 π ⇒ | tan( y x )| ≤ 1 4 π ⇒ −π
4 ≤ tan( y x ) ≤ π 4
⇒ arctan( −π
4 ) ≤ y x ≤ arctan(π 4 ) ⇒ −1 ≤ y x ≤ 1.
Angular region of angle π 2
symmetric to the positive x-axis is found.
9. If Rez ≤ Imz and z = x + iy, then x ≤ y. So
10. For Re( 1 z
) < 1, we take z = x + iy. Because we know that
we conclude that
1
Re(
x + iy ) ⇒
1
z = 1
x + iy =
This is the exterior of the circle of radius 1 2 centered at 1 2 .
x
x 2 + y 2 ,
x
x 2 + y 2 < 1 ⇒ 1 4 < (x − 1 2 )2 + y 2 .
11