Complex Number Inequality

Prove that if

\left|z + 4\right| + \left|z - 4\right| = 10

, then

\left|z\right| \geq 3

where

z

is a complex number.

I tried proving this for a long time but didn't succeed. My textbook provides a very counter-intuitive proof. Can anyone please help me with this?

Reply 1

zetamcfc

Original post by esrever

Prove that if

\left|z + 4\right| + \left|z - 4\right| = 10

, then

\left|z\right| \geq 3

where

z

is a complex number.

I tried proving this for a long time but didn't succeed. My textbook provides a very counter-intuitive proof. Can anyone please help me with this?

Think what this equation represents on an Argand diagram. Then you should see why the statement is true.

Reply 2

pibobelebeep

I’m currently studying for a level math and I’m not very good at it, but If I did this I think I would so something like this
3B069527-5D55-49BA-9CF3-90FAE0CF44E2.jpg.jpeg

3B069527-5D55-49BA-9CF3-90FAE0CF44E2.jpg.jpeg

Sorry for messy handwriting

Reply 3

esrever

Original post by zetamcfc

Think what this equation represents on an Argand diagram. Then you should see why the statement is true.

Thanks! I tried plotting it. I think it represents an ellipse which does prove the inequality.

But can you help me prove it algebraically?

Reply 4

esrever

Original post by pibobelebeep

I’m currently studying for a level math and I’m not very good at it, but If I did this I think I would so something like this
3B069527-5D55-49BA-9CF3-90FAE0CF44E2.jpg.jpeg

Sorry for messy handwriting

Thanks. But I think there is a flaw in your proof.

|z + 4| \neq \sqrt{z^2 + 4^2}

Reply 5

zetamcfc

Original post by esrever

Thanks! I tried plotting it. I think it represents an ellipse which does prove the inequality.

But can you help me prove it algebraically?

We think about what |z|=>3 means? The distance from the origin to any complex number on the ellipse must be greater than or equal to 3. So we find the closest point on the ellipse to the origin to find the smallest value |z| can take.

(edited 4 years ago)

Reply 6

pibobelebeep

Whoops, sorry, well I guess I better start practising more. hehehe...
and I don't know if you already saw this but https://math.stackexchange.com/questions/2551813/modulus-of-complex-number-and-inequality had a really similar question.

Original post by esrever

Thanks. But I think there is a flaw in your proof.

|z + 4| \neq \sqrt{z^2 + 4^2}

Reply 7

esrever

Original post by zetamcfc

Thank you so much for the response. But doesn't this assume that |z| is smallest where the ellipse cuts the x-axis. I can see why that is intuitively true but I'm having trouble proving it. Anyways, is there a way to prove the theorem without using the intuition from argand diagrams?

Reply 8

zetamcfc

Original post by esrever

The closest point is where the ellipse cuts the imaginary axis. It is the closest point as a circle of that radius will be ''inscribed'' in the ellipse. Now as you know it is purely imaginary, you get that z=bi where b is a real number.

Now you can solve for b, in the new equation, |bi-4|+|bi+4|=10.

This question is all about understanding what you are actually doing and what you are being asked to show.