Complex Numbers VI: Finding Least Value of |z-2√2-4i|

In summary: Type |z-4i|<=\sqrt{5} to get |z-4i|<=\sqrt{5}. In summary, the problem is to find the least value of |z-2\sqrt{2}-4i| over the intersection of the two loci defined by |z-4i|<=\sqrt{5} and \frac{\pi}{4}<=arg(z+4)<=\frac{\pi}{2}. To find z_1, we need to find the shortest distance from a point P to a semi-disk. To do so, we use the formula \frac{|a\overline{b}-b\overline{a}|}{|a-b|} where a
  • #1
Punch
44
0
Sketch on an Argand diagram the set of points satisfying both |z-4i|<=\sqrt{5} and \frac{\pi}{4}<=arg(z+4)<=\frac{\pi}{2}.

I have already sketched the 2 loci. The problem lies in the following part.

Hence find the least value of |z-2\sqrt{2}-4i|. Find, in exact form, the complex number z_1 represented by the point P that gives this least value.
 
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  • #2
Punch said:
Hence find the least value of |z-2\sqrt{2}-4i|.
Over which set of z? The intersection defined in the first part? It seems that you need to find the shortest distance from a point to a semi-disk.

Why don't you wrap the [tex]...[/tex] tags around your formulas?
 
  • #3
Evgeny.Makarov said:
Over which set of z? The intersection defined in the first part? It seems that you need to find the shortest distance from a point to a semi-disk.

Why don't you wrap the \(z_1\) tags around your formulas?

Yes, how do I then find the complex number [tex]z_1[/tex] in the following part?

I tried using the latex but they didnt seem to work
 
  • #4
Punch said:
Yes, how do I then find the complex number [tex]z_1[/tex] in the following part?
See the following picture.

argand.png


Punch said:
I tried using the latex but they didnt seem to work
Type [tex]\frac{\pi}{4}\le\arg(z+4)\le\frac{\pi}{2}[/tex] to get [tex]\frac{\pi}{4}\le\arg(z+4)\le\frac{\pi}{2}[/tex].
 
  • #5


To find the least value of |z-2\sqrt{2}-4i|, we need to find the closest point to the origin on the locus |z-2\sqrt{2}-4i|. This can be done by setting z=x+yi and using the distance formula.

Distance formula: d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Substituting z=x+yi and the given values of 2√2 and 4i, we get:

d=\sqrt{(x-2√2)^2+(y-4)^2}

Since we want to find the closest point to the origin, we can set x=0 and y=0:

d=\sqrt{(0-2√2)^2+(0-4)^2}

d=\sqrt{4+16}=√20=2√5

Therefore, the least value of |z-2√2-4i| is 2√5.

To find the exact complex number z_1 represented by the point P that gives this least value, we can substitute the values of x and y into z=x+yi:

z_1=0+0i=0

Therefore, the complex number z_1 is 0.

On the Argand diagram, this point would lie on the x-axis, 2√2 units to the left of the origin. This satisfies both loci and gives the least value of |z-2√2-4i|.

 

Related to Complex Numbers VI: Finding Least Value of |z-2√2-4i|

1. What is the purpose of finding the least value of |z-2√2-4i|?

The least value of |z-2√2-4i| helps to determine the minimum distance between the complex number z and the point (2√2, 4) on the complex plane.

2. How do you find the least value of |z-2√2-4i|?

To find the least value of |z-2√2-4i|, we can use the distance formula d = √((x2-x1)^2 + (y2-y1)^2). In this case, x1 = 2√2 and y1 = 4. We can plug in the x and y values of any complex number z and calculate the distance to (2√2, 4). The minimum distance will be the least value of |z-2√2-4i|.

3. Can the least value of |z-2√2-4i| be a negative number?

No, the least value of |z-2√2-4i| cannot be a negative number. The absolute value of a complex number is always a positive number, so the least value will also be positive.

4. What does the least value of |z-2√2-4i| tell us about the location of z on the complex plane?

The least value of |z-2√2-4i| tells us the minimum distance between z and the point (2√2, 4) on the complex plane. If the least value is 0, then z is located at (2√2, 4). If the least value is greater than 0, then z is located outside of the point (2√2, 4).

5. How can the concept of the least value of |z-2√2-4i| be applied in practical situations?

The concept of the least value of |z-2√2-4i| can be applied in various fields such as engineering and physics. For example, it can be used to calculate the minimum distance between two points in a circuit or the minimum distance between a moving object and a fixed point. It can also be used in optimization problems to find the minimum distance between two objects or the minimum error in a measurement.

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