Need help with a question on complex number

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January 19th 2011, 06:09 PM
leonheart1234

Need help with a question on complex number

It says to find 4 different complex number z1,z2,z3,z4 such that each complex number through these 4 properties will produce an integer.
property 1) Re(z) = integer
property 2)Im(z) = integer
property 3) Re(z^(-1)) = integer
property 4)Im(z^(-1)) = integer
where Re and Im denotes the real and imaginary of the complex number respectively.

Please help.
I came up with only 1, i need 3 more.
The one i came up with is
z1=1+1i
January 19th 2011, 06:11 PM
tonio

Quote:

Originally Posted by leonheart1234

It says to find 4 different complex number z1,z2,z3,z4 such that each complex number through these 4 properties will produce an integer.
property 1) Re(z) = integer
property 2)Im(z) = integer
property 3) Re(z^(-1)) = integer
property 4)Im(z^(-1)) = integer
where Re and Im denotes the real and imaginary of the complex number respectively.

Please help.
I came up with only 1, i need 3 more.
The one i came up with is
z1=1+1i

This question was already solved some 1 week ago.

Tonio
January 19th 2011, 06:28 PM
leonheart1234

Yes,I know its similar.But the notation is different.
In the other post it says Re(z-1) and Im(z-1)
Mine is Re(z^(-1)) and Im(z^(-1)).
January 20th 2011, 04:16 AM
tonio

Quote:

Originally Posted by leonheart1234

Yes,I know its similar.But the notation is different.
In the other post it says Re(z-1) and Im(z-1)
Mine is Re(z^(-1)) and Im(z^(-1)).

Well, it's pretty similar, and if you did the past question and understood it then

it should be clear what to do in this case.

If $z=x+iy\in\mathbb{C}$ , what is then $z^{-1}$ ? After you answer this then try your problem.

Tonio
January 20th 2011, 04:31 AM
Plato

Quote:

Originally Posted by leonheart1234

It says to find 4 different complex number z1,z2,z3,z4 such that each complex number through these 4 properties will produce an integer.
property 1) Re(z) = integer property 2)Im(z) = integer
property 3) Re(z^(-1)) = integer property 4)Im(z^(-1)) = integer

Try: $z_1=1,~z_2=i,~z_3=-1,~\&~z_4=-i~.$

Why do they work?
January 20th 2011, 05:59 PM
Krahl

let $z = x + iy$
then $z^{-1} = \frac{1}{z} = \frac{1}{x + iy}= \frac{1}{x + iy} * \frac{x - iy}{x - iy}= \frac{x - iy}{x^2 + y^2}= \frac{x}{x^2 + y^2} - i\frac{y}{x^2 + y^2}$
then $Re(z^{-1}) = \frac{x}{x^2 + y^2}$ ----------- (1)
and
$Im(z^{-1}) = \frac{-y}{x^2 + y^2}$ --------------- (2)

You need to find integers x and y such that (1) and (2) are also integers. Your solution of z1=1+1i with x = 1 and y = 1 is incorrect since (1) gives $\frac{1}{2}$ .

In fact you can go further and find all the possible solutions to your question.

For x and y integers, $x^2 + y^2 = k^2 \geq 1$ must also be an integer. This equation is a circle centered about the origin and only has 4 solutions (along the x and y axes). x = k, y = 0 and x= -k, y=0 etc...

for $|x| > 1$ or $|y| > 1$ , $x^2 + y^2 >|x|$ or $x^2 + y^2 > |y|$ so one of (1) or (2) will be a fraction. So
$|x| = 1$ or $|y|=1$ .

HallsofIvy: either |x|= 1 and y= 0 or |y|= 1 and x= 0. Thus giving the four solutions Archie Moore gave.
January 21st 2011, 04:59 AM
HallsofIvy

Quote:

Originally Posted by Krahl

let $z = x + iy$
then $z^{-1} = \frac{1}{z} = \frac{1}{x + iy}= \frac{1}{x + iy} * \frac{x - iy}{x - iy}= \frac{x - iy}{x^2 + y^2}= \frac{x}{x^2 + y^2} - i\frac{y}{x^2 + y^2}$
then $Re(z^{-1}) = \frac{x}{x^2 + y^2}$ ----------- (1)
and
$Im(z^{-1}) = \frac{-y}{x^2 + y^2}$ --------------- (2)

You need to find integers x and y such that (1) and (2) are also integers. Your solution of z1=1+1i with x = 1 and y = 1 is incorrect since (1) gives $\frac{1}{2}$ .

In fact you can go further and find all the possible solutions to your question.

For x and y integers, $x^2 + y^2 = k^2 \geq 1$ must also be an integer. This equation is a circle centered about the origin and only has 4 solutions (along the x and y axes). x = k, y = 0 and x= -k, y=0 etc...

for $|x| > 1$ or $|y| > 1$ , $x^2 + y^2 >|x|$ or $x^2 + y^2 > |y|$ so one of (1) or (2) will be a fraction. So
$|x| = 1$ or $|y|=1$ .

You should say a bit more: either |x|= 1 and y= 0 or |y|= 1 and x= 0. Thus giving the four solutions Archie Moore gave.