locus (2 Viewers)

[scora]

1. find the locus of z, given t is a real variable:

z = (2+it) / (2-it)


2. prove that:

(|z|- iz) / (|z|+ iz) = -i(sec0+tan0) , where Re(z)is not equal to zero, and argz=0

NOTE: [0=theta]
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[lolokay]

for the second one

z = x + iy
z* = conjugate of z = x - iy
zz* = |z|² = x² + y²

(|z|- iz) / (|z|+ iz) * (|z| - iz*)/(|z| - iz)*
= (|z|² - zz* - i|z|(z+z*)/(|z|² + zz* + i(z-z*))
= -i(2x)/(2|z| + i2iy)
= -i(x)/(|z| - y) * (|z| + y)/(|z| + y)
= -i(x|z| + xy)/(x² + y² - y²)
= -i(|z|/x + y/x)
= -i(sec@ + tan@)
since x=|z|cos@, and y/x=tan@

 

[Trebla]

1. find the locus of z, given t is a real variable:

z = (2+it) / (2-it)

z = (2 + it) / (2 - it)
= (2 + it)² / (4 + t²)
= (4 - t² + 4it) / (4 + t²)
Let z = x + iy
x + iy = (4 - t² + 4it) / (4 + t²)
Equating real and imaginary parts:
x = (4 - t²) / (4 + t²)
y = 4t / (4 + t²)
We have a parametric equation with parameter t. To find the Cartesian equation, we must eliminate t.
4x + xt² = 4 - t²
t²(x + 1) = 4 - 4x
t² = 4(1 - x) / (1 + x)
t = ± 2 √[(1 - x) / (1 + x)] (with x =/= -1)

Case 1 (t > 0): Sub positive case into y
y = 8√[(1 - x) / (1 + x)] / [4 + 4(1 - x) / (1 + x))
= 2√[(1 - x) / (1 + x)] / [1 + (1 - x) / (1 + x)]
= 2√[(1 - x) / (1 + x)] / [(1 + x + 1 - x) / (1 + x)]
= 2√[(1 - x) / (1 + x)] / [2 / (1 + x)]
= √[(1 - x) (1 + x)]
= √(1 - x²) (with x =/= -1)

Case 2 (t < 0): Sub negative case into y
y= - 8√ [(1 - x) / (1 + x)] / [4 + 4(1 - x) / (1 + x))
= - √(1 - x²) similarly (with x =/= -1)

Therefore, the locus is y = ±√(1 - x²)
i.e. x² + y² = 1 (excluding the point (-1,0))

Alternatively (I just realised this after all that algebra lol), you could use a geometric interpretation. The division of a complex number and its conjugate leads to division of their moduli and subtraction of their arguments.
So z = (2 + it) / (2 - it) = √(4 + t²) cis θ₁ / √(4 + t²) cis θ₂
= cis (θ₁ - θ₂)
Notice the moduli are equal (both are √(4 + t²)), so when they divide they cancel to 1.
This means the modulus of z is 1.
Also, θ₂ = - θ₁ as the conjugates have equal and opposite arguments
So z = cis (2θ₁)
Now the argument 2θ₁ varies according to t. So for all values of t, you have all possible values for an argument except that 2 + it and 2 - it cannot be purely imaginary i.e. θ₁ =/= π/2 => 2θ₁ =/= π
The result is a unit circle excluding the point (-1, 0) as expected.

 

[gurmies]

Nice work Trebla, I found the algebraic way somewhat more difficult. The geometric way occurred to me far quicker.

 

[lolokay]

"So for all values of t, you have all possible values for an argument."

θ1 =/= ±pi/2, so 2θ =/= pi, so you don't have -all- possible values (the point (-1,0) is missing from the unit circle).
you can see from your algebraic solution that x=/=-1

 

[Trebla]

Ah yes, well spotted...forgot about that