triangle inequality for complex numbers

[illjay7005]

Is there a way to verify the triangle inequality for complex numbers without using the conjugate, and just using the definition of modulus?

So I need to show that |z1 + z2| <= |z1| + |z2|

I am starting with just |z1 + z2|=|(a+bi)+(c+di)|= |(a+c) +(b+d)i|= [ (a+c)^2 + (b+d)^2 ]^1/2 by definition of modulus

With just algebraic manipulation, I have gotten this to [ (a+b)^2 + (c+d)^2 + 2(ac +bd) ]^1/2

This is close to what I need but I know I cannot break up this square root, and I have this extra term.
This is where I am stuck, because the only properties that I see from here involve the conjugates. I know that 2(ac + bd) is 2Re(z1z2'), but this uses conjugates. I have the property |Re(z)| <= |z| but I am not seeing how I can use that because I cannot break up the square root. What am I missing here? Any suggestions? I appreciate any help. Thanks!!

 

[DrMike]

Is there a way to verify the triangle inequality for complex numbers without using the conjugate, and just using the definition of modulus?

So I need to show that |z1 + z2| <= |z1| + |z2|

I am starting with just |z1 + z2|=|(a+bi)+(c+di)|= |(a+c) +(b+d)i|= [ (a+c)^2 + (b+d)^2 ]^1/2 by definition of modulus

With just algebraic manipulation, I have gotten this to [ (a+b)^2 + (c+d)^2 + 2(ac +bd) ]^1/2

This is close to what I need but I know I cannot break up this square root, and I have this extra term.
This is where I am stuck, because the only properties that I see from here involve the conjugates. I know that 2(ac + bd) is 2Re(z1z2'), but this uses conjugates. I have the property |Re(z)| <= |z| but I am not seeing how I can use that because I cannot break up the square root. What am I missing here? Any suggestions? I appreciate any help. Thanks!!

You'll probably find it easier to prove |z1+z2|^2 <= (|z1|+|z2|)^2 than to prove directly |z1+z2| <= |z1|+|z2|.