# Vector subtraction

• Nov 19th 2012, 10:28 PM
Tutu
Vector subtraction
Hi I really need help here.

I'm getting conflicting answers for both questions..

For the image with the P and Q complex number, to find the distance PQ, the textbook says to take z-w.

However, for the image of a parallelogram, to find PR, it is w-z.

I am really confused, If they ask to find eg. AB in terms of a and b, shouldn't B be the place where both arrowheads meet? And thus, A will be -A and thus, AB = b-a?!

Can someone explain what went wrong with my thinking process, I really am confused.

Thank you so much,
J.
• Nov 19th 2012, 10:45 PM
Deveno
Re: Vector subtraction
let's say we start at point A. from point A, we have 2 vectors, one going FROM point A TO point B (direction matters), and one going from point A to point C.

now if we want to know the vector going FROM B to C, we have the following situation:

vector AB = u
vector AC = v

vector AB + vector BC = vector AC (we vector sum by going "head to tail").

that is:

u + vector BC = v

vector BC = v - u.

if we want to find vector CB (start from C, go TO B), we have:

vector AC + vector CB = vector AB

v + vector CB = u

vector CB = u - v.

comparing the two, we see vector BC = -vector CB (same vector, but "opposite sign").

in the picture on the left, we want to find the vector from P to Q. this is what we know:

z + PQ = w (both z + PQ and w start at the origin, and end up at Q).

therefore PQ = w - z.

so why does your book say "z -w"?

because if all we want is the DISTANCE, we take |w - z| = |z - w| (sign doesn't matter for distance).
• Nov 19th 2012, 11:09 PM
Tutu
Re: Vector subtraction
Thank you!

My book really says z-w as attached.
Attachment 25818

If we are not looking for the distance, then whether it is z-w and w-z does matter doesn't it? If z-w gives -4+8i, then w-z will give 4-8i, both vectors are different in terms of direction but equal in magnitude. But since they are not on the same plane, they are different vectors, is this right? Please correct me if I'm wrong, I'm not good at this!

Thank you so much,
J.
• Nov 19th 2012, 11:59 PM
Deveno
Re: Vector subtraction
yes they are negatives of each other. for any vector, v, v and -v lie on the same LINE (if we picture them as starting "at the origin"), but the "direction of travel" is opposite.

your book is correct (barely), but ONLY because it is finding "distance". your confusion is understandable, and it reflects poorly on your text that they are not more "consistent".
• Nov 20th 2012, 12:23 AM
Tutu
Re: Vector subtraction
Thank you!

By the way, can I clarify one more doubt about the 'distance' part. I understand it but with regards to another question,

If z is a+bi,
|z| is sqrt(a^2 + b^2)
If I'm looking at -iz,
-iz will be b-ai,
|-iz| is sqrt((b)^2+(-a)^2) = sqrt(a^2 + b^2)

With both, I end up with sqrt(a^2 + b^2), does this mean that the distance between the two complex number z and -iz on the Argand diagram are the same? Which doesn't really make sense somehow..
What would make sense to me is that it would mean the distance from the y-axis to each complex number is the same. But I've never heard of that..
What would be it mean, for z and -iz to have sqrt(a^2 + b^2) which is supposedly distance? How do I interpret it?

Thank you so very much really,
J.
• Nov 20th 2012, 02:20 AM
Deveno
Re: Vector subtraction
no the distance between z and the origin, and -iz and the origin are the same.

how do we get -iz from z? we multiply by -i (in a bizarre twist of fate, the additive and multiplicative inverses of i are the same: 1/i = -i).

what does multiplication by -i do? it rotates a vector z by 90 degrees clockwise. a rotation around the origin does not change a points distance from the origin.

for example z-->-iz takes z = 1 to -iz = -i. that is, it takes the vector (1,0) to (0,-1).

perhaps you are used to seeing rotations as a 2x2 matrix. let's prove that the matrix R:

[ 0 1]
[-1 0] does the same thing as multiplying by -i:

what is R(a,b)? it's (b,-a), so R takes a+bi to b-ai.

and (-i)(a+bi) = -ia - bi2 = -ia - (b)(-1) = -ia + b = b-ai.

sometimes complex multiplication is called "rotation-dilation". if we write z = |z|(cos(arg(z)) + i sin(arg(z))).

then the map w-->zw takes |w|(cos(arg(w)) + i sin(arg(w))) to |z||w|(cos(arg(z)+arg(w)) + i sin(arg(z)+arg(w)))

so "multiplying by z" rotates w by the angle arg(z) and stretches (dilates) the magnitude by a factor of |z|.

the complex number -i can be written as:

-i = |-i|(cos(arg(-i)) + i(sin(arg(-i))) = cos(arg(-i)) + i sin(arg(-i))) (since -i has magnitude 1)

= cos(-π/2) + i sin(-π/2) (and of course, cos(-π/2) = cos(π/2) = 0, and sin(-π/2) = -sin(π/2) = -1, so -i = 0 + (-1)i, no big surprise).

***********

this explains one reason why for real numbers: negative*negative = positive.

multiplying by a negative number once "rotates 180 degrees" (in this case it doesn't matter if it's +180 or -180, you wind up at the same point).

if you do this twice, you've rotated by 360 degrees in all, so you're back where you started (direction-wise).

it also explains why we put i and -i on the y-axis:

a square root of -1 should be "halfway" (rotation-wise) from 1 to -1, that is: on the y-axis (90 degrees, + for counter-clockwise, - for clockwise (by convention)), so that when we "rotate twice", we've gone a total of 180 degrees.
• Nov 20th 2012, 04:56 AM
Tutu
Re: Vector subtraction
Thank you, I really appreciate this.

But this is how I thought of it, to which had me thinking that distance between z and the origin, and -iz and the origin are not the same."

If
z = a+bi
z is a units right and b units up from the origin ie. second quadrant.
Yet,
-iz is b-ai
thus -iz is b units right and a units down from the origin ie. fourth quadrant.

Since they are different units ( one is a to the right, the other is b to the right etc. ) up and down from the origin, how is it then that their distances from the origin is the same?

Thank you very much for your time,
J.
• Nov 20th 2012, 06:23 AM
Plato
Re: Vector subtraction
Quote:

Originally Posted by Tutu
If z = a+bi
z is a units right and b units up from the origin ie. second quadrant.
Yet, -iz is b-ai
thus -iz is b units right and a units down from the origin ie. fourth quadrant.
Since they are different units ( one is a to the right, the other is b to the right etc. ) up and down from the origin, how is it then that their distances from the origin is the same?

Do you understand that if each of $\displaystyle z~\&~w$ is a complex number then it is true that $\displaystyle |z\cdot w|=|z|\cdot|w|$?

Well $\displaystyle |iz|=|i||z|=(1)|z|=|z|$.

All of these are true.
$\displaystyle |z|=|-z|=|\overline{z}|=|iz|=|-iz|$
• Nov 20th 2012, 01:34 PM
Deveno
Re: Vector subtraction
think of it like this:

if you have a right triangle with vertical leg a and horizontal leg b, it has the same length hypotenuse as a right triangle with horizontal leg a and vertical leg b.

(because the hypotenuse has length√(a2+b2), which is the same as√(b2+a2)).

moreover, if you flip either of these triangles "upside down", or rotate them, the hypotenuse is STILL the same length.

so replacing one (or both) a or b by it's negative, still doesn't change the length of the hypotenuse, because we SQUARE the distance of the legs before adding them.

in fact, all complex numbers z with the same distance from the origin form a CIRCLE:

the set {z: |z| = r > 0} is a circle in the complex plane. this is more easily seen by writing z = x + iy, where we get the set:

{z = (x,y): √(x2 + y2) = r > 0}. since x2+yy and r are both positive, we lose no information by squaring both sides to get the set:

{(x,y): x2 + y2 = r2}, which is the USUAL equation defining a circle.

that is what a circle IS: a ring of points all the same distance from the circle's center. and |z| just encodes what the distance from 0+0i to z is, different complex numbers z = a+bi and z' = a'+b'i can still have the same distance from 0 as long as:

a2+b2 = a'2+b'2.

for example, 1 and i are both "distance one away from 0" even though 1 has the coordinates (1,0) and i has the coordinates (0,1).

the complex number √2/2 + (√2/2)i is also "one away from 0" because:

(√2/2)2 + (√2/2)2 = 2/4 + 2/4 = 4/4 = 1.