Question

How to demostrate this? (complex numbers)

Answer 1

z=x+iy

\[\left| x \right|+\left| y \right|\le \sqrt{2}\left| z \right|\]

Answer 2

I've can't find how to do this

Answer 3

the |...| sign on both sides means magnitude, right ?

so,
|z| = \(\sqrt{x^2+y^2}\)

and |x| = x, |y| = y because they are real numbers

Answer 4

yep

Answer 5

so we just need
x+y < = sqrt 2 sqrt (x^2+y^2)

Answer 6

that is
\(\sqrt {x^2+y^2} \le \sqrt {2x^2+2y^2}\)

or
we need to prove
x^2+y^2 <= 2x^2 +2y^2

Answer 7

got this ?

Answer 8

no, wait

Answer 9

I think I got it

Answer 10

\(\sqrt {x^2+2xy + y^2} \le \sqrt {2x^2+2y^2}\)

Answer 11

that ^

Answer 12

which is easy to prove.....
you got it this way only ?

Answer 13

Well...with drawings

Answer 14

Why is the left side that?

Answer 15

the |x| + |y|
???

magnitude of real numbers is itself...
|x| = x , |y| = y

x+y = \(\sqrt {(x+y)^2}\)
....

Answer 16

oh right

Answer 17

One last question...how can I find a solution to this?

\[\left( z+1 \right)^{2}=3+4i\]

Answer 18

expand left side
first take
z = x+ iy
so,
[(x+1) + iy]^2
expand this
then compare the real and imaginary parts

Answer 19

you get something like
(x+1)^2+ y^2 = 3
2(x+1)y = 4

Answer 20

2 equations 2 unknowns, see if you can solve it.....doesn't seem easy

Answer 21

negative :)

Answer 22

-y^2 in first one, right?

Answer 23

ya, right, -y^2

Answer 24

okk...let's try the system

Answer 25

y =2/ (x+1)
put this in 1st equation...

Answer 26

I got x3+3x2=4...is it correct until there?

Answer 27

Nope, just found a mistake

Answer 28

:/

Answer 29

that was because i used +y^2 again :P
anyways,
i get integer solutions now
one solution we can easily guess by trying x=1

Answer 30

How did you get that?

Answer 31

i'll tell u,
just tell me what equation u got in x ?

Answer 32

What I did was x2(x+1)+2x(z+1)-2=2x+2

Answer 33

and then tried to solve for that

Answer 34

how u got that equation ?

Answer 35

maybe go step by step from starting....so that i can identify whether u made any error or not..

Answer 36

I got that (x+1)^2-y2=3

Answer 37

I plug in y=2/(x+1)

Answer 38

oh, so u didn't square y ...

Answer 39

y^2 = 4/ (x+1)^2
plug this in

Answer 40

(x+1)^2 + 4/ (x+1)^2 = 3

Answer 41

maybe to simplify things, you can assume (x+1)^2 = a
first and get a quadratic in a

Answer 42

a+4/a =3

Answer 43

why did you plug in that?

Answer 44

why did i plug in a ?

Answer 45

It is 2(x+1)y=4 at the start?

Answer 46

yeah, so ?

Answer 47

oh...I think I get it now

Answer 48

good, try to solve further
i took 'a' bcoz,
if i would have worked with x , i would have got x^4 +... = 0
which is not easy to solve
so i took (x+1)^2 =a

find a, then resubstitute it by (x+1)^2
then find x

Answer 49

ok, I will do it :) thanks

Answer 50

tag me if you get stuck
or if u want to verify your final answer...

Answer 51

sure :)