complex Q (1 Viewer)

[vds700]

R is a positive real number and z₁ , z₂ are complex numbers. Show that the points on the Argand
diagram which represent respectively the numbers z₁ , z₂ ,
(z₁ - iRz₂)/(1 - iR)

form the vertices of a right angled triangle.

 

[u-borat]

wait uh....1 is a complex number?

 

[vds700]

wait uh....1 is a complex number?

sorry, typo, fixed up now

 

[u-borat]

wow man that's a pretty hard question unless im missing something.
tried to realise the denominator, that only made the question harder. -_-

 

[Iruka]

R is a positive real number and z₁ , z₂ are complex numbers. Show that the points on the Argand
diagram which represent respectively the numbers z₁ , z₂ ,
(z₁ - iRz₂)/(1 - iR)

form the vertices of a right angled triangle.

Call that sicko number w.

We need to use the converse of Pythagoras' theorem, that is, we want to show that

|z₁ - z₂|^2 = |z₁-w|^2 + |z₂-w|^2

Now to do this, you need to remember that is |z|^2 = z times the conjugate of z. If you do that, the algebra is not actually all that bad - you'll see that the denominators sort themselves out into something relatively friendly.

 

[conics2008]

hey Vds Im sorry im just oging to ask a complex numbe rrealted question..

given that u(w) + (u)w = where (w) and (u) is its conj..

prove that Real(u(w)) = 0 ???

I hate algebra based complex numbers.

I trie dusing the top one but i dont know where to go on from there.

 

[conics2008]

uh?

its the conjucate

 

[vds700]

helops if u post the question correctly

This is the way my tutor showed me how to do this Q a while ago, not sure about its validity.

Typed this up in paint lol

 

[shinn]

Or alternatively:

let:
u = a+bi
(u) = a - bi
w = x + yi
(w) = x - yi


Since (u)w + (w)u = 0

Therefore,
(a-bi)(x+yi) + (x-yi)(a+bi) = 0
Expanding and simplifying:
2(ax+yb)
ax + yb = 0


But,
u(w) = (a+bi)(x-yi) = (ax +by) + (xb-ay)i = (xb-ay)i
Thus Re(u(w)) = 0



For the previous question:
Let A represent z1, B represent (z1 - iRz2)/(1 - iR) and C represent z2.

Let @ represent the angle ABC

@ = arg (AB>) - arg (CB>) (AB> and CB> are vectors)
= arg ( (z1 - iRz2)/(1 - iR) - z1 ) - arg ( (z1 - iRz2)/(1 - iR) - z2)
= arg (z1 - z2) - arg ( (z1-z2)iR ) , (after making common denominator & simplifying)
= arg ( (z1 - z2) / ((z1-z2)iR))
= arg (1/iR)
= arg ((-1/R)i)
= pi/2

Thus, angle ABC is right angled and triangle ABC must represent vertices of a right angled triangle.

 

[lmaosine]

 

[friction]

Im so screwed for complex. Back to the textbook for me.

 

[u-borat]

helops if u post the question correctly

This is the way my tutor showed me how to do this Q a while ago, not sure about its validity.

Typed this up in paint lol

yeah that's valid thats how i would have done it.