Triangle Inequality of Complex Number

Triangle Inequality of Complex Number states that the absolute value of sum of two complex number is always less than or equal to the sum of individual absolute value of the complex numbers. Absolute value of the complex number represents here the modulus value of the complex number. It is very important concept of complex number algebra.
If we have complex number z & w, then triangle inequality states that

(abs.Z + abs. W ) is greater than abs, Z+W

So many times in the course of learning complex number we have to use this triangle inequality of modulus of complex number.

I believe this Triangle Inequality of Complex Number tutorial was helpful to enhance your concept in Algebra Mathematics.

[jiqbal]

wow! you have really have talent in explaining this. I will hopefully watch more of your videos. plus i really like your accent, its cool. i wish you luck

[headfmob]

Thanks alot for this explanation, the other method

|z + w|2 = (z + w)·(z + w)' is very tough to digest

thanks again bro :)

[ItsMyAcademy]

oo wooo., .. thanks brother for inspiring.. you like my accent, people say is funny, some people its like old 70s movie.. and bla bla ... thanks good luck to you too.. :)

[ItsMyAcademy]

You are very welcome brother... You can digest brother...ha ha tell me how can I help you ?

[goodcallyall]

at 3:22, you square both sides of your inequality in red. But (ac + bd) could be negative, so surely this is a flawed step?

[sahihaibaap]

is there any shorter method with less calculation..

[tini]

well as far as i think it is pretty easy but it based on the statement that we hae to prove so why don't u use this approach |z|=z.conjugate of z

[vfp]

Your proof is flawed, I'm afraid. As @goodcallyall points out "at 3:22, you square both sides of your inequality in red. But (ac + bd) could be negative", so you can't square it (-1 < 0 but (-1)^2 > 0). You can fix it easily enough though.
Multiply the inside of the square roots and you get
Without loss of generality, assume |c|>|a| and |d|>|b|, (this is the trick)
which gives you
<
Now, = (c^4+2(cd)^2+d^4)^(1/2) = (c^2+d^2). This gets you
(ac-bd)<(c^2+d^2)
0<c^2+d^2ac-bd
0<c^2-ac+d^2-bd
0<c(c-a)+d(d-b)
Since we assumed |c|>|a| and |d|>|b|, the above is true. There is no loss of generality, because whichever c or a, or d or b, then the product of each will be positive.

[jaefarshameem]

cant we just draw the vector diagram it literally takes like 2 seconds

[shrimatkapoor]

i love his English it's so shit and good at the same time

[gurman]

Well proof never starts LHS and RHS taken simultaneously!

[jerkmanjesus]

This doesnt show anything about the graphical method of using the triangle inequality.

Put another way. No triangles were pictured in this film, so not useful.