Complex Q (1 Viewer)

[azureus88]

Find the locus of z such that Im[(z+1)/(z+i)]=Re[(z+1)/(z+i)]

The answer's meant 2 be x^2 + y^2 = 1, but im getting (x+1)^2 + (y+1)^2 =1.

Can someone confirm the right answer plz?

 

[Trebla]

Let z = x + iy
(z + 1)/(z + i) = (x + 1 + iy) / (x + i(y + 1))
= (x + 1 + iy)(x - i(y + 1)) / (x² + (y + 1)²)
= (x(x + 1) - i(x + 1)(y + 1) + ixy - i²y(y + 1)) / (x² + (y + 1)²)
= (x² + x + y² + y + i(- xy - x - y - 1 + xy)) / (x² + (y + 1)²)
= (x² + x + y² + y + i(- x - y - 1)) / (x² + (y + 1)²)
lm[(z + 1)/(z + i)] = Re[(z + 1)/(z + i)]
=> (x² + x + y² + y) / (x² + (y + 1)²) = (- x - y - 1) / (x² + (y + 1)²)
x² + x + y² + y = - x - y - 1
x² + 2x + y² + 2y + 1 = 0
x² + 2x + 1 + y² + 2y + 1 = 1
(x + 1)² + (y + 1)² = 1
Yeah, unless my working is wrong, I got your answer.
And looking more closely, I think I know where the mistake occured.
It used (x² + x + y² + y - i(x + y + 1)) / (x² + (y + 1)²) and ignored the negative on the i when it came to finding real and imaginary parts which ended up with:
x² + x + y² + y = x + y + 1
=> x² + y² = 1

 

[azureus88]

btw, how do u find the cartesian equation to the locus of arg[(z+1)/(z+i)]=(pi/3or4pi/3), using geometric approach?

 

[Trebla]

To find the Cartesian equation, you simply have to locate the centre and calculate the radius of a full circle and apply restrictions on that equation which equate to the arc. This requires using the fact that the angle at the centre of a circle is twice that of the angle at the circumference, then use a bit of trigonometry and coordinate geometry to locate the centre and find the radius. Though it would pretty rare for you to be required to do that...

 

[azureus88]

hey, i got another question. arent the square roots of [(root3)+i] meant to be (root2)(cis(pi/12)) and (root2)(cis(13pi/12))? apparently its +or-(root2)(cis(pi/12)).

can someone explain this plz?

 

[tommykins]

sqrt3+i = 2cis(pi/6)
z^2 = 2cis(pi/6)
thus z = +-sqrt2cis(pi/12)