Finding complex number with the lowest argument.

In summary, you solved the problem by finding the point S with the lowest possible argument, which is (20,0). You then computed the arcsin function from the triangle POS, getting the result (1,-1).
  • #1
cdummie
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5

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.
 
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  • #2
cdummie said:
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}
##
 
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  • #3
Samy_A said:
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.
 
  • #4
cdummie said:
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.
 
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  • #5
A geometrical approach would also work, but I don't want to distract you if you don't need another approach.
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##.
 
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  • #6
Samy_A said:
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?
 
  • #7
cdummie said:
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.
 
  • #8
Samy_A said:
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.
 
  • #9
cdummie said:
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##.

Snapshot.jpg

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.:smile:)
 
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  • #10
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.
 
  • #11
Deleted as irrelevant.
 
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1. What is the definition of a complex number?

A complex number is a number that can be written in the form a + bi, where a and b are real numbers and i is the imaginary unit, equal to the square root of -1.

2. How do you find the argument of a complex number?

The argument of a complex number can be found by taking the inverse tangent of its imaginary part divided by its real part. It is also known as the angle or phase of the complex number in the complex plane.

3. Can a complex number have a negative argument?

Yes, a complex number can have a negative argument. This means that the complex number is located in the fourth quadrant of the complex plane, where both the real and imaginary parts are negative.

4. What is the significance of finding a complex number with the lowest argument?

Finding a complex number with the lowest argument is important in applications such as signal processing and electrical engineering. It helps to determine the phase difference between two signals and can be used to analyze the stability of a system.

5. How do you find the complex number with the lowest argument?

To find the complex number with the lowest argument, you can use the polar form of a complex number, which is z = r(cos θ + i sin θ). The argument of the complex number is represented by θ, and the complex number with the lowest argument will have the smallest value of θ.

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