Finding complex number with the lowest argument.

[cdummie]

Homework Statement

Of all complex numbers that fit requirement: $|z-25i| \leq 15$ find the one with the lowest argument.

Homework Equations

The Attempt at a Solution

z=a + ib (a, b are real numbers)

$\sqrt{a^2 + (b-25)^2} \leq 15 \\ a^2 + (b-25)^2 \leq 225$

The lowest possible argument is zero, in that case, complex number has only real part, which means that imaginary part is zero, it is obvious that none of the numbers with such argument fit this requirement, because, if b is zero then on the LHS we have a squared plus 625, which is way greater than 225 even without a^2.

 

[Samy_A]

z=a + ib (a, b are real numbers)

$\sqrt{a^2 + (b-25)^2} \leq 15 \\ a^2 + (b-25)^2 \leq 225$

You have to bring the argument $\phi$ into play.
You could transform your inequality by using:
$\begin{cases} a=r\cos \phi \\ b=r\sin \phi \end{cases}$

 

[cdummie]

You have to bring the argument $\phi$ into play.
You could transform your inequality by using:
$\begin{cases} a=r\cos \phi \\ b=r\sin \phi \end{cases}$

Ok, let's see,

$r^2cos^2\phi + (rsin\phi - 25)^2 \leq 225 \\ r^2cos^2\phi + r^2sin^2\phi - 50rsin\phi + 625 \leq 225 \\ r^2 -r50sin\phi +400 \leq 0$

So i ended up with quadratic equation by unknown r, but i don't know $\phi$ either.

 

[Samy_A]

Ok, let's see,

$r^2cos^2\phi + (rsin\phi - 25)^2 \leq 225 \\ r^2cos^2\phi + r^2sin^2\phi - 50rsin\phi + 625 \leq 225 \\ r^2 -r50sin\phi +400 \leq 0$

So i ended up with quadratic equation by unknown r, but i don't know $\phi$ either.

Rewrite the inequality with $\phi$ on one side and r on the other side:
$\frac{r²+400}{50r} \leq \sin \phi$
Now remember you have to find the lowest possible $\phi$.
The LHS is clearly positive, so you are looking for the r-value that minimizes that LHS.

 

[Samy_A]

A geometrical approach would also work, but I don't want to distract you if you don't need another approach.

Spoiler

Take the tangent from the origin to the circle centered at $25i$ with radius 15 (the tangent with positive slope, as we want the lowest positive argument). The point you need is the intersection of the tangent and the circle.
This will easily give you r, and, from there you get $\phi$.

 

[cdummie]

Rewrite the inequality with $\phi$ on one side and r on the other side:
$\frac{r²+400}{50r} \leq \sin \phi$
Now remember you have to find the lowest possible $\phi$.
The LHS is clearly positive, so you are looking for the r-value that minimizes that LHS.

But, how can i find the lowest possible value, is it by taking derivative?

 

[Samy_A]

But, how can i find the lowest possible value, is it by taking derivative?

Yes, find a zero of the derivative and (if there is one) establish that it is a minimum.

 

[cdummie]

Yes, find a zero of the derivative and (if there is one) establish that it is a minimum.

I think i solved it, lowest r = 20 so $\phi = \arcsin\frac{4}{5}$.

But, i would really like if you could explain me geometrical approach (i hope I am not bothering you too much), it might help me understand what is really going on there.

 

[Samy_A]

I think i solved it, lowest r = 20 so $\phi = \arcsin\frac{4}{5}$.

But, i would really like if you could explain me geometrical approach (i hope I am not bothering you too much), it might help me understand what is really going on there.

So $z=12 + 16i$.

You are looking for the point $z$ that satisfies $|z-25i| \leq 15$ and with the lowest possible argument.
$z$ has to be in (or on) the circle with centre at (0,25) and with radius 15.
You will get the lowest argument with the tangent from the origin to the circle. That gives you the point S (or $z$ as complex number). As POS is a right triangle, you easily get that r=20.
And then you can compute $\arcsin \phi$ from the same triangle POS.

(Sorry for the clumsy drawing, never used that whiteboard before. P is supposed to be the centre of the circle.)

 

[Mark44]

Keep in mind that |z - 25i| represents the distance from a complex number z to the complex number 25i. You should be able to solve this problem simply by drawing a diagram.

 

[Samy_A]

Deleted as irrelevant.