Use z = a + ib, w = c+id and the triangle inequality to prove that
√[(a+c)² + (b+d)²] ≤ √(a² + b²) + √(c² + d²)

Question

Use z = a + ib, w = c+id and the triangle inequality to prove that 
√[(a+c)² + (b+d)²] ≤ √(a² + b²) + √(c² + d²)

dumbcow · Answer

square both sides and simplify
$$ ac+bd \le \sqrt{a ^{2}+b ^{2}}\sqrt{c ^{2}+d ^{2}}$$
square both sides and simplify
$$2abcd \le b ^{2}c ^{2} + a ^{2}d ^{2}$$
make substitution
u =bc
v = ad
$$2uv \le u ^{2}+v ^{2}$$
look at all 3 cases, 
u=v
2u^2 <= 2u^2
True

u>v:  u=v+1
$$2v(v+1)\le v ^{2}+(v+1)^{2}$$
$$\rightarrow 2v ^{2}+2v \le 2v ^{2}+2v+1$$
0 <= 1
True


u<v :  u=v-1
$$\[2v(v-1)\le v ^{2}+(v-1)^{2}$$ $$\rightarrow 2v ^{2}-2v \le 2v ^{2}-2v+1$$
0 <= 1
True

anonymous · Answer

Ignore the method above; it is not what they want.


Consider |z| and |w|.

We can consider these as the distances from the origin to each complex number in an argand diagram.

Now consider |z - w|. What does this mean? This is:
$$|z-w| = \sqrt{(a-c)^2 + (b-d)^2} $$
so the distance between the two complex numbers, z and w.

However, this is not going to be exactly what we want for the LHS, so let us instead consider the complex numbers z and -w (and |z-(-w)|.

We work out the values of each of these things:

$$ |z| = \sqrt{a^2 + b^2}$$
$$ |-w| = \sqrt{(-c)^2 + (-d)^2} = \sqrt{c^2 + d^2}$$
$$ |z-(-w)| = |z+w| = \sqrt{(a+c)^2 + (b+d)^2}$$

But how can we use this to prove the inequality? Wait, DUH, we use the triangle inequality.

Considering a triangle with sides a,b,c, we can say:
$$a \leq b+c$$ with the same holding for each side in a's position.

How can this relate to our problem? We know |z| and |-w| are the distances from the origin to each of these, and |z+w| is the distance between them. OWAIT these form a triangle. From this we can say:

$$|z+w| \leq |z| + |w| $$

and the result follows immediately.


-----

Sorry for the long post, the method is actually very quick but as you aren't here for me to guide through it I thought I'd explain everything - this method is FAR better than above (and what they actually wanted)

Peace out, INewton.

anonymous · Answer

I'll be back later to collect my medals.

anonymous · Answer

lol

anonymous · Answer

here

anonymous · Answer

Unfortunately, you gave one to the first post too, so that is unacceptable.

anonymous · Answer

I thought you were harsh.

anonymous · Answer

Thank you very much! I thought that was a bit harsh too, but the first answer confused me. Yours was very helpful and the answer that I was looking for. Medal given ^^