complex number

Subhotosh Khan

Super Moderator
Staff member
View attachment 29652

May I ask some help on how to do part a ?
but I don't get the logic and would really some explanation that leads to that.
TIA
I do not see any 'z' in the figure you have posted.

Zermelo

Junior Member
View attachment 29652

May I ask some help on how to do part a ?
but I don't get the logic and would really some explanation that leads to that.
TIA
What I presume is that the equation in a) holds true for every complex number z on the circle. What you have to understand is that |a-b| represents the distance between two complex numbers a and b. Also, a circle with circumference r and center in the point O is the set of all points that are at an r distance from the point O (center of the circle, think of how a compass works!). In this picture, try to find the center of your circle, it will be a complex number (point) at the imaginary axis. Also, try to see what is the circumference of your circle. Then, for every point (complex number) z on the circle, |z-O| = r will hold!

• Subhotosh Khan

nanaseailie

Junior Member
I do not see any 'z' in the figure you have posted.
z=a+bi (it's like x for any variable), but I share the same confusion too with you....

nanaseailie

Junior Member
What I presume is that the equation in a) holds true for every complex number z on the circle. What you have to understand is that |a-b| represents the distance between two complex numbers a and b. Also, a circle with circumference r and center in the point O is the set of all points that are at an r distance from the point O (center of the circle, think of how a compass works!). In this picture, try to find the center of your circle, it will be a complex number (point) at the imaginary axis. Also, try to see what is the circumference of your circle. Then, for every point (complex number) z on the circle, |z-O| = r will hold!
thank you for the explanation, i get some idea, but why is it |z+2i|=2 as an answer, not |z-2i|=2.
may I also ask why the formula for circle that you wrote is |z-O| = r, not |z+O| = r.
I am still confused with the sign, but I'm getting there!

nanaseailie

Junior Member
does it have any connection with the circle equation (x-a)^2 + (y-b)^2 = r^2 ?

• Zermelo

Dr.Peterson

Elite Member
thank you for the explanation, i get some idea, but why is it |z+2i|=2 as an answer, not |z-2i|=2.
may I also ask why the formula for circle that you wrote is |z-O| = r, not |z+O| = r.
I am still confused with the sign, but I'm getting there!
Just as the distance between two points a and x on the number line (as in your equation of a circle) is |x-a|, the complex-number "distance" (or vector) from O to Z is Z-O, and the actual distance is |Z-O|.

Here, the center, O, is the point (0,-2), or the complex number -2i; and z - (-2i) = z + 2i.
does it have any connection with the circle equation (x-a)^2 + (y-b)^2 = r^2 ?
Yes, it's the same basic idea, and you could even use this form directly, though working with the complex numbers directly is simpler in some ways. You'd have z = (x,y) and O = (a,b) = (0,-2).

• nanaseailie

nanaseailie

Junior Member
Just as the distance between two points a and x on the number line (as in your equation of a circle) is |x-a|, the complex-number "distance" (or vector) from O to Z is Z-O, and the actual distance is |Z-O|.

Here, the center, O, is the point (0,-2), or the complex number -2i; and z - (-2i) = z + 2i.

Yes, it's the same basic idea, and you could even use this form directly, though working with the complex numbers directly is simpler in some ways. You'd have z = (x,y) and O = (a,b) = (0,-2).
My Gosh thank you Dr. Peterson,
your explanation gives some light! I would like to say deeply thank you for your kind time and explanation!

• Dr.Peterson