Need some help

[Daenerys Targaryen]

Can someone help me through the workings of this...?
Let z=x+yi and w=u+vi
Prove that:
 |z+w| is less than or equal to |z| + |w|

Cheers !

[Homer]

 |z+w| is less than or equal to |z| + |w|
I get upto



but im stuck here :/

[brightsky]

note to homer: (xv - yu)^2 always greater or equal to 0.

[polar]

Spoiler

other
1. why is true:

,
thus, is positive and is positive, while in the expression , and aren't necessarily positive, hence the second one is always either greater or equal to the first one.

2. thanks battman for the \overline thing

[nerdgasm]

|z+w| is less than or equal to |z| + |w|
I get upto



but im stuck here :/

I think you're very nearly there, what brightsky (and I) did was notice that
, therefore
, and transposing the equation gives the desired inequality.

Here's my try at the full algebraic proof (although the one by polar is certainly more elegant)

Spoiler

We have and
Therefore, and

Now, we square both terms, to obtain
, and


Here, I subtracted one expression from the other, leading to

                                           

Now, we aim to show that (as this proves )

Let and
Note that is always non-negative, so if . We will now analyse the case where

Squaring both and we obtain



Now, , and using the inequality provided at the beginning of this post, it is proven that , therefore and hence

Putting it all together, we have shown that , and hence , hence .

Making the obvious rearrangement, we obtain

Finally, notice that and are always non-negative, therefore, we may take the positive square root of both sides of the inequality above, to obtain
, as shown.

There is also a geometric argument that can be used to show this inequality.
Using an Argand diagram, draw a vector starting at the origin and ending at to denote the complex number . You can pick the magnitude and direction of this vector.

Now, draw a second vector starting at in the exact same direction as the first vector. You can make this second vector as long as you like. The magnitude of this second vector is .

You effectively have one long vector going out from the centre of the Argand diagram. Draw a circle using this long vector as a radius. This large circle has radius

Now draw a second circle, centred at , with the radius of the second vector. This smaller circle shows all the possible results of adding to a complex number of magnitude (you determined the magnitude when originally drawing ).

Notice that your second circle is completely contained within the first (it meets the first at one point, and 'curves' away more quickly than the first). Therefore, as this second circle shows all the possible values of for a given and , and is completely contained within a circle of radius , it follows that , as shown.

[Homer]

haha i get it, thanks guys suppose i get stuck on the last step as shown above in the exam, would i get the majority of the marks, since i've done all the working out but haven't done the last step? :/