We have just been introdued to the fascinating manipulations of Complex Numbers and hence I know the basics only. Please give me the solution without reference to higher concepts to the problem.
Prove that
|z₁ + z₂ | < = |z₁ | + |z₂ |
where z_i E C and |a| represents the magnitude / modulus of the complex number a.

I'll start off by giving a horrible algebraic proof. At the end I'll give you a hint about the geometrical method, which actually shows intuitively why the inequality is true.

Define:

z₁ = x₁ + iy₁
z₂ = x₂ + iy₂

where x₁ ,x₂ ,y₁ ,y₂ are real then:

|z₁ + z₂ | = |(x₁ + x₂ ) + i(y₁ + y₂ )|
= sqrt((x₁ + x₂ )² + (y₁ + y₂ )² )

and

|z₁ | = sqrt(x₁ ² + y₁ ² )
|z₂ | = sqrt(x₂ ² + y₂ ² )

So we have to show:

sqrt((x₁ + x₂ )² + (y₁ + y₂ )² ) < = sqrt(x₁ ² + y₁ ² ) + sqrt(x₂ ² + y₂ ² ) (*)

This can be done as follows. We know:

(y₁ x₂ - x₁ y₂ )² > = 0

Expanding:

y₁ ² x₂ ² + x₁ ² y₂ ² > = 2x₁ y₁ x₂ y₂

Add x₁ ² x₂ ² + y₁ ² y₂ ² to each side and factorise:

(x₁ ² + y₁ ² )(x₂ ² + y₂ ² ) > = (x₁ x₂ + y₁ y₂ )²

Since each factor on the right hand side is positive we can take the square root:

sqrt(x₁ ² + y₁ ² )sqrt(x₂ ² + y₂ ² ) > = x₁ x₂ + y₁ y₂

Double this inequality, add x₁ ² + x₂ ² + y₁ ² + y₂ ² to each side and then factorise:

(sqrt(x₁ ² + y₁ ² ) + sqrt(x₂ ² + y₂ ² ))² > = (x₁ + x₂ )² + (y₁ + y₂ )²

and (*) then follows upon taking square root.

If you want an intuitive idea for why the inequality is true, try drawing the points represented by z₁ , z₂ and z₁ + z₂ on an Argand diagram. Then use the fact that |z| means the length of the line between 0 and z on the Argand diagram (for any complex number z) and the required inequality follows by geometry.

I already know the geometric method, I was looking for the purely algebraic one !
It is just a simple application of the relations (inequalities ) between the sides of a triangle and is therefore called the triangle inequality. But I wanted to have the purely algebraic proof as I was interested.