Locus - Complex numbers (1 Viewer)

[YBK]

Hey, just a question.

Sketch on an Argand diagram the locus of the point P representing z, given that |z|^2 = z + complex conjugate of z + 1

Thanks!

 

[Mountain.Dew]

Hey, just a question.

Sketch on an Argand diagram the locus of the point P representing z, given that |z|^2 = z + complex conjugate of z + 1

Thanks!

use the substitution method --> z = x + iy

so, you would obtain --> |x+iy|^2 = x+iy + (x +1 - iy)

x^2 + y^2 = 2x + 1

x^2 - 2x + y^2 = 1

x^2 - 2x + 1 + y^2 = 2

(x-1)^2 + y^2 = 2 ==> locus is a circle, radius sqrt 2, centre (1,0)

 

[YBK]

use the substitution method --> z = x + iy

so, you would obtain --> |x+iy|^2 = x+iy + (x +1 - iy)

x^2 + y^2 = 2x + 1

x^2 - 2x + y^2 = 1

x^2 - 2x + 1 + y^2 = 2

(x-1)^2 + y^2 = 2 ==> locus is a circle, radius sqrt 2, centre (1,0)

How did you get the 2nd step.

(x + iy)^2 = x^2 - y^2 + 2xiy

or did u equate the real parts... but in that case, shouldn't it be x^2 - y^2 and not x^2 + y^2 ?

 

[Riviet]

How did you get the 2nd step.

(x + iy)^2 = x^2 - y^2 + 2xiy

or did u equate the real parts... but in that case, shouldn't it be x^2 - y^2 and not x^2 + y^2 ?

It's not (x + iy)², it's |x + iy|²! Pay attention to if they're brackets or absolute value symbols.
So |x + iy|²= [sqrt(x²+y²)]²=x²+y²
Also, notice how the imaginary part of z and conjugate of z cancel out, ie (x+iy)+(x-iy)=2x.
If you have any more problems, feel free to ask.

 

[YBK]

It's not (x + iy)², it's |x + iy|²! Pay attention to if they're brackets or absolute value symbols.
So |x + iy|²= [sqrt(x²+y²)]²=x²+y²
Also, notice how the imaginary part of z and conjugate of z cancel out, ie (x+iy)+(x-iy)=2x.
If you have any more problems, feel free to ask.

ohhh

just remembered |z| is the modulus! And the modulus is root x^2 + y^2 therefore x^2 + y^2 is the modulus squared!

Thanks riv and mountain dew!

 

[Riviet]

No problem, glad I helped.

 

[Mountain.Dew]

a pleasure YBK, a pleasure