Finding Integral Re/Im Parts of Complex Numbers

[SeannyBoi71]

Homework Statement

Find four complex numbers z each with the property that Re(z), Im(z), Re(z^-1), Im(z^-1) are all integers, where Re and I am denote the real and imaginary parts respectively of a complex number.

Homework Equations

Maybe 1/z = [itex]\frac{\bar{z}}{|z|²}[/itex] ? On my screen that code isn't working... hopefully you guys can see it.

The Attempt at a Solution

I'm not sure if I have to use the above equation, or the answer is as simple as the numbers being (1+i), (-1-i), (1-i), (-1+i). Because the real and imaginary parts of each complex number will be an integer, as well as when they are reciprocated. If they were anything other than "1" they wouldn't be an integer...

 

[cepheid]

Homework Equations

Maybe 1/z = [itex]\frac{\bar{z}}{|z|²}[/itex] ? On my screen that code isn't working... hopefully you guys can see it.

Don't use the [noparse] [/noparse] tags for superscripts within the tex tags. The former are just for superscripting plain text on the forums. The proper way to achieve supercripts and subscripts in LaTeX is to use ^ and _ respectively. See for example the following:

$$\frac{1}{z} = \frac{\bar{z}}{|z|^2}$$

(right click on the equation and select "Show Source" from the menu that comes up in order to see how I did that.

I think the problem with your reasoning is that it's NOT true that if z = a + ib, then 1/z = (1/a) + i(1/b)

Instead, the reciprocal of z is another number w such that zw = 1. Since it's true that zz* = |z|^2, the equation you posted for 1/z is the right thing to pursue. (Note, I use the * notation for the complex conjugate instead of a bar).

 

[SeannyBoi71]

Ah, thanks for clearing that up.

Alright, is this the only way to go about the question? Our prof never went over this equation, I just found it in the textbook. Most of the time he doesn't make us use a technique we didn't practice in class, but this may be different.

 

[SeannyBoi71]

Instead, the reciprocal of z is another number w such that zw = 1. Since it's true that zz* = |z|^2, the equation you posted for 1/z is the right thing to pursue. (Note, I use the * notation for the complex conjugate instead of a bar).

I'm just going to go ahead and try the question using that equation. I just attempted it and basically plugged in $$z = (a+bi)$$ and came up with a whole mess.

Edit: that was a really stupid question. Anyhow, that mess is $$\frac{1}{z} = \frac{a^3-ab^2-ba^2i+b^3i}{a^4-b^4}$$ Is this right, and where do I go from here?

 

[HallsofIvy]

No, that is not right! Where in the world did you get it? If z= a+ bi then
$$\frac{1}{z}= \frac{1}{a+ bi}= \frac{1}{a+ bi}\frac{a- bi}{a- b}= \frac{a- bi}{a^2+ b^2}$$

 

[SammyS]

So a²+b² has to divide a , & a²+b² has to divide b .

It looks like a or b has to be zero.

 

[SeannyBoi71]

No, that is not right! Where in the world did you get it? If z= a+ bi then
$$\frac{1}{z}= \frac{1}{a+ bi}= \frac{1}{a+ bi}\frac{a- bi}{a- b}= \frac{a- bi}{a^2+ b^2}$$

Alright, well aren't you supposed to multiply the top and bottom by the conjugate of a+bi? In the denominator you have it multiplied by a-b and in the numerator it's a-bi...

 

[SammyS]

Alright, well aren't you supposed to multiply the top and bottom by the conjugate of a+bi? In the denominator you have it multiplied by a-b and in the numerator it's a-bi...

So, if you want this to be purely real, what must b be equal to?

 

[SeannyBoi71]

So a²+b² has to divide a , & a²+b² has to divide b .

It looks like a or b has to be zero.

So I figure HallsofIvy's response had a typo in it, anyways...

I tried 4 scenarios: a=0 and b=1 gives -i, a=1 and b=0 gives 1, a=0 and b=-1 gives i, and a=-1 and b=0 gives 1. Two of the answers of the same so this cannot be the final answer. What can I do about th3 4th number? Thanks for leading me through this guys, appreciate it...

 

[SeannyBoi71]

So, if you want this to be purely real, what must b be equal to?

It must be 0 so that 0*i=0, right? I'm still a little confused...

 

[SammyS]

All four of those are different.

 

[SeannyBoi71]

Guess that's true. Just overlooked it. Many thanks