Let \(z\) and \(w\) be complex numbers such that \(|z + 1 + 3i| = 1\) and \(|w - 7 - 8i| = 3.\) Find the smallest possible value of \(|z - w|.\)

 Sep 9, 2022

Hello WorldEndSymphony!

We need to know the geometric meaning of the following:  (Also, are "hints").

1-     \(\left | z-(a+bi) \right |=r\) is a circle with centre (a,b) and radius r.

2-     \(\left |z-w \right |\) is the distance between z and w.

And, to solve this question, we need to sketch the givens.


Rewrite:  \(\left | z+1+3i \right |=1 \iff \left |z-(-1-3i) \right |=1\)

Sketch this in an Argand diagram.

Next: \(\left | w-7-8i\right |=3 \iff \left | w-(7+8i) \right |=3\)

Sketch this also in the same Argand diagram.

We should get:

But you might be wondering, where is z and w? 
What we have drawn is a "Locus" you see, "z" can be any point on the red circle (on that red line only as it is "=" , if for example it were 

\( \left |z-(-1-3i) \right |<1\)   "<" then any point inside the circle might be z, and if  ">" then outside the circle.)

Similarly, for "w", it is any point on the blue circle (That blue lines only). 

So what the question is asking, what is the smallest distance between these circles such that the distance is between their circumferences.


The shortest distance must be perpendicular to the tangent at the point to any circle right? (As I drawn it, this must be the shortest distance.) But, since d is perpendicular to both tangents there, then it if we extended d, it must pass through both centers of the circles.

(A more rigourous proof of this can be obtained maybe from youtube or google search.) But, intuitively at least it makes sense!

But, again, think about how to find "d". 

Hint: we know the radius of each circle!
Hint: distance formula!

Ok, we will find the distance between the two centers: \(\sqrt{(7+1)^2+(8+3)^2}=\sqrt{185}\)

Now, we remove the two radii to get:

\(d=\sqrt{185}-1-3=\sqrt{185}-4\) which is what we wanted :).


I hope this helps!



 Sep 10, 2022

wooow!!! what a slick solution :D thank you so much for helping me!!!!

WorldEndSymphony  Sep 11, 2022

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