# Using the triangle inequality, prove for any complex number z, that |Re(z)|+|Im(z)|<= sqrt(2)*|z|I'm really not sure how to do this. Here's what I've done so far. Let z=a+bi, such that a and b...

Using the triangle inequality, prove for any complex number z, that |Re(z)|+|Im(z)|<= sqrt(2)*|z|

I'm really not sure how to do this. Here's what I've done so far.

Let z=a+bi, such that a and b are real.

Consider the square of |a| + |b|

This is a^2+ 2*|a|*|b| + b^2.

Since |a|<|z| and b<|z|, my expression above is less than or equal to

a^2 + b^2 + 2*|z|^2.

Since |z|^2=a^2+b^2, the above is equal to

|z|^2+2*|z|^2

Thus 3*|z|^2 >= (|a|+|b|)^2.

Thus sqrt(3)*|z| >= |a|+|b|

I'm not sure how to use the triangle inequality, or get the coefficient of |z| to be sqrt(2). Help please!

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### 3 Answers

Like the answer above, I can show this without using the triangle inequality.

Suppose we consider the complex number `z=a+ib` as a right-angled triangle, with legs `a=Re(z)` and `b=Im(z)` , and hypotenuse `r=|z|=sqrt{a^2+b^2}` .

Now we also know that for any positive numbers `a` and `b` , that the perfect squares `(a+b)^2` and `(a-b)^2` are both positive or zero. Starting with the second one, we have:

`(a-b)^2>=0` expand

`a^2+b^2-2ab>=0` move term to right side

`a^2+b^2>=2ab` now add `a^2` and `b^2` to both sides

`a^2+b^2+a^2+b^2>=2ab+a^2+b^2` simplify each side

`2(a^2+b^2)>=(a+b)^2` read inequality right-to-left instead of left-to-right

`(a+b)^2<=2(a^2+b^2)` replace right side with definition

`(a+b)^2<=2|z|^2` take square root of both sides. Keep only positive root

`a+b<=sqrt 2|z|` replace with definitions

`|Re(z)|+|Im(z)|<sqrt2 |z|` and it has been solved

**The inequality has been proved.**

Well, I can show you where the `sqrt(2)` comes from, and I can prove it in a way similar to your attempt, but I can't find a nice way to involve the triangle inequality. So here's a sort of half answer to your question.

If we use polar form and write `z=rcostheta+irsintheta` , so that `Re(z)=rcostheta`, `Im(z)=rsintheta` , and `|z|=|r|`, the given inequality is equivalent to `|rcostheta|+|rsintheta|<=sqrt(2)|r|`, or factoring out the `r` and dividing,

`|costheta|+|sintheta|<=sqrt(2)`. Now to prove this, since both sides are non-negative, square both sides to get the equivalent

`cos^2x+sin^2x+2|cosxsinx|<=2`, or `|2cosxsinx|<=1`. But `|2cosxsinx|=|sin(2x)|<=1`, which completes the proof.

I'd imagine that a proof involving the triangle inequality might involve some clever way of adding zero, such as `|Re(z)+Im(z)|=|Re(z)+iIm(z)-iIm(z)+Imz|`. Hopefully someone else can show us the way, and I'll keep working at it.

Another related solution is to just directly use the Cauchy-Schwarz (CS) Inequality, from which the result follows almost immediately.

`|Re(z)|+|Im(z)|=1*|Re(z)|+1*|Im(z)|<=sqrt(1^2+1^2)sqrt(|Re(z)|^2+|Im(z)|^2)`

`=sqrt(2)|z|` ,

where the inequality is true by CS. Just another example of the useful "multiply by 1" trick.