triangular inequalities (complex) (1 Viewer)

[jnney]

If |z1|=6 and z2=4+3i, show that the greatest value of |z1+z2| is 11 and find its least value.

 

[bleakarcher]

|z1+z2|<=|z1|+|z2| (Triangle Inequality)
i.e. |z1+z2|<=6+5=11
Hence, max |z1+z1|=11.
Picture:

Sorry dude, I dont know about the second part.

 

[RealiseNothing]

khfreakau's got this

 

[khfreakau]

I got the lowest value possible for |z1 + z2| is the square root of 109.

EDIT: I forgot to consider negatives, will have another attempt.

My new answer is square root of (61-42root2) which is approximately 1.266.

this is wrong, you can easily get a modulus of 1 by letting arg(z1)=arg(z2)+pi, and even then lesser moduli are possible. i'm working on it now.

edit: the question is effectively "find the least modulus of (x-4)^2 + (y-3)^2 = 36".

edit2: wait no, i made a calculation error. the least modulus is one.

to solve it, you could simply deduce that the locus that satisfies the two given conditions for z1 and z2 means that z1 must lie on a circle of radius 6, and z2 is a fixed point. hence the locus is a circle of radius six with centre z2.

from here, you could either simply do it graphically, or use the arg thing i mentioned earlier.

by letting x=0 and y=0, you can see if the intercepts will have a lesser value, which they do not (you can deduce this graphically) and so yeah, the least modulus is 1.

 

[RealiseNothing]

this is wrong, you can easily get a modulus of 1 by letting arg(z1)=arg(z2)+pi, and even then lesser moduli are possible. i'm working on it now.

edit: the question is effectively "find the least modulus of (x-4)^2 + (y-3)^2 = 36".

edit2: wait no, i made a calculation error. the least modulus is one.

z2 is in the first quadrant, so arg(z2) + pi would be in the second quadrant.

When you add these complex numbers together you're going to get a modulus greater than 1 aren't you?

z1 would have to be either in the third of fourth quadrant so that you get the smallest modulus possible? So letting arg(z1) = arg(z2) + pi wouldn't give the smallest possible modulus.

Could you explain how you got 1 as an answer, I want to see what I'm doing wrong.

 

[khfreakau]

z2 is in the first quadrant, so arg(z2) + pi would be in the second quadrant.

When you add these complex numbers together you're going to get a modulus greater than 1 aren't you?

z1 would have to be either in the third of fourth quadrant so that you get the smallest modulus possible? So letting arg(z1) = arg(z2) + pi wouldn't give the smallest possible modulus.

Could you explain how you got 1 as an answer, I want to see what I'm doing wrong.

sure. when you add pi, you won't get an argument in the second quadrant, you'll get an argument in third quadrant. any argument + pi will result in an argument in the opposite quadrant, ie if the original argument was in quadrant 1, then the resultant one would be in q3. if it was in q2, it would then be in q4 and so on.

effectively, if you add pi to an argument, and plot the lines that result, you'll get a straight line, and this will point in the opposite direction. this means that when you add the complex numbers the modulus will be effectively "cancelled" since you have a modulus of 6 in the opposite direction, and a fixed modulus of 5, so you'll get a complex number that has an arg of tan^-1(3/4) +- pi (adding or subtracting pi is the same thing in terms of the principal argument of complex numbers), and a modulus of one.

if that doesn't make sense i can draw up a diagram if you like

 

[RealiseNothing]

sure. when you add pi, you won't get an argument in the second quadrant, you'll get an argument in third quadrant. any argument + pi will result in an argument in the opposite quadrant, ie if the original argument was in quadrant 1, then the resultant one would be in q3. if it was in q2, it would then be in q4 and so on.

effectively, if you add pi to an argument, and plot the lines that result, you'll get a straight line, and this will point in the opposite direction. this means that when you add the complex numbers the modulus will be effectively "cancelled" since you have a modulus of 6 in the opposite direction, and a fixed modulus of 5, so you'll get a complex number that has an arg of tan^-1(3/4) +- pi (adding or subtracting pi is the same thing in terms of the principal argument of complex numbers), and a modulus of one.

if that doesn't make sense i can draw up a diagram if you like

strictly speaking it would have made more sense for me to say that we should let arg(z2)=arg(z1) +- pi, such that arg(z2) lies in the third quadrant.

Thanks, I know what I was doing wrong now lol. I confused 'pi' with 'i' and thought that adding pi would add 90 degrees(which is what you do when multiplying by i).

So simple and obvious now.

 

[math man]

there are two forms of the triangle inequality which you frequently use to find min and max mod of |z1 + z2|... the first form which everyone should know is : |z1 + z2| <= |z1| + |z2| which geometrically
means that the third side of a triangle is less than the sum of the other two sides and equality occurs when all three vectors are collinear.... the second form which not everyone is too familiar is:
|z1 + z2| => |z1| - |z2| or |z1+z2| => |z2| - |z1|, the order depends on whether |z1| or |z2| is bigger and it is obvious which form should be used as |z1+z2| should always be positive...equality here
occurs when the second vector is the negative of the first times some integer...and geometrically it reads that the third side of a triangle is greater than the difference of the other two...now combing these
two gives: |z1|-|z2| <= |z1 + z2| <= |z1| + |z2| and so we use this to determine the max and min of |z1+z2| as this inequality is always true for all |z1| and |z2|...so in the above example the max is
simple found by the addition of the moduli, i.e 5 + 6 = 11, and the least value obviously will have to be 6-5= 1...hopes this helps a bit..

 

[tohriffic]

Hi I solved this geometrically.

Note that B represents (4 + 3i) or in other words z2.

With pythagoras' theroem, AB is 5. So BC is 1 (min value) and BD is equal to 11 (max value)

I hope you can gather all that from the info. on the diagram. If not, feel free to ask. ^ ^

Just remember, you don't have to solve every question algebraically.