Question

I have a question from my textbook.

If s and t are two complex number,
and
(a) (s*t)*=st*
(b) st*+s*t is a real number, and st*-s*t is an imaginary number.
(c) (st*+s*t)^2-(st*-s*t)^2= (2|st|)^2
(d) (|s|+|t|)^2 - |s+t|^2 = 2|st| - (st*+s*t)

prove that |s+t|<= |s|+|t| (algebraically)

Answer 1

What does (s*t)* mean?
Typically * means to multiply

Answer 2

* means the conjugate of the complex number

Answer 3

you are asked to prove abc and d (which i've been able to do it myself) and use these results to prove |s+t|<= |s|+|t| which I don't have any idea how to go about it.

Answer 4

So it is really, prove that:
\[\sqrt{(a+c)^{2}+(b+d)^{2}}\le \sqrt{a ^{2}+b ^{2}}+\sqrt{c ^{2}+d ^{2}}\]

Answer 5

here is an idea, although it may take a bunch of boring computation. you have
\[(|s|+|t|)^2 - |s+t|^2 = 2|st| - (s\overline{t}+\overline{s}t)\] so it looks like from the way you were walked through this that the point is to show that the right hand side of the equality is positive.
not that i did it, but it looks like that is what you have been lead to

Answer 6

I don't know how I should go about introducing an inequality sign into the equation and unfortunately the text book don't provide answers for questions involving proof...

Answer 7

@Mertsj in a nutshell.. yes. There also has been a mention on the equation that is to be proved is called triangle inequalities.

Answer 8

Here's one way to do this.

Answer 9

thx. been bugging me since 2 days ago.