Posted by **Cambridge** on Thursday, January 2, 2014 at 11:24am.

Prove algebraically that |w+z| is less than or equal to |w|+|z| for any complex numbers w and z, where || is the magnitude.

After letting w = a+bi and z = c + di and doing some plugging in etc. I got that

ac-bd <= b^2 + sqrt (a^2+b^2)(c^2+d^2) + d^2

But is my proof complete? I know logically sqrt (a^2+b^2)(c^2+d^2) is a bit less than ac+ad+bc+bd but now what?

## Answer This Question

## Related Questions

- Complex Numbers - Let z and w be complex numbers such that |z| = |w| = 1, and zw...
- Complex Numbers - Let z and w be complex numbers such that |z| = |w| = 1, and zw...
- Precal - Prove algebraically that |w+z|<=|w|+|z| for any complex numbers w ...
- math-complex numbers - Sketch the sets of complex numbers in the complex plane ...
- Pre-cal - The equation of the line joining the complex numbers -5 + 4i and 7 + ...
- calculus - can you please explain tep by step how to solve dividing complex ...
- Algebra proof - Prove that if w,z are complex numbers such that |w|=|z|=1 and ...
- Algebra II - 9x+y is less than or equal to 1700 9x+3y is less than or equal to ...
- Precalculus - Let z and w be complex numbers such that |z| = |w| = 1, and zw is ...
- Math - Let z and w be complex numbers such that |z| = |w| = 1, and zw is not ...

More Related Questions