complex numbers question.....plz help???????

Question

if I z-1 I <3, then prove that I iz +3 -5i I < 8

Answer 1

First off, let us consider the following inequality:
|z1+z2| <= |z1|+|z2|,
where <= represents 'less than or equal to'. Now, consider
|z-1|<3.
Let us replace z-1 by a new complex number, say, w. Then, we have:
|w| < 3.

Consider what we have to prove:
|iz+3 -5i|.
We may write this as:
|i(z-1) + 3 - 4i| = |wi + a|,
where a is the complex number |3-4i|. Now, apply the inequality:
|wi + a| <= |wi| + |a|, where |a| =5 and |wi| = |w|. Thus:
|wi + a| <= |w| + 5. But, |w| < 3, which is the strict inequality given to us, thus we conclude that
|wi + a| <= |w| + 5 < 3+5 < 8!

QED!

Answer 2

You can't say -2 < z < 4 because z is a complex number and therefore it is not restricted to a range of reals. If someone knows that their knowledge in a topic is limited then why do they contribute?

I can't write the lines either side so I will use mod to represent modulus of a complex number.

mod[z - 1] < 3 means that z can be any point on an Argand diagram inside a circle of radius 3 with centre (1,0).
The transformation iz moves this to any point within a circle of radius 3 with centre (0,1) on the Argand diagram or 0 + i.
The further transformation + 3 - 5i means that it is now a point inside a circle radius 3 with centre (3,-4) or 3 - 4i. This final centre is 5 units from the origin. Therefore any point inside this circle cannot be more than 5 + radius = 8 units from the origin.
Therefore mod[iz +3 - 5i] < 8.

Answer 3

if I z-1 I < 3
z - 1 < 3
z < 4
- z + 1 < 3
- z < 2
z > 2,
2 < z < 3
I iz +3 -5i I < 8 has a slightly different meaning. It means the magnitude of iz + 3 - i5 < 8 or
3^2 + (z - 5)^2 < 64
Using z = 2,
9 + (2 - 5)^2 = 18
Using z = 3,
9 + (3 - 5)^2 = 13
13 and 18 are both less than 64.