Imaginary numbers and circles

kmalik001

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Feb 10, 2013
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Hey so I have another complex numbers question. My main issue is I am getting confused with the modulus sign. In this case how would I get rid of it? Square both sides. If I square the modulus what would be the answer like z^2 what? Once I get past this I am confident I can solve the question.

Sketch the circle described by the equation

|z − u| = 2
Determine the points of intersection (if any) of this circle and the two axes, real and imaginary.

where u = −1 + j√3
 
I know this modulus question is very basic and I am embarassed to ask it but i really couldn't understand online/
 
Hey so I have another complex numbers question. My main issue is I am getting confused with the modulus sign. In this case how would I get rid of it? Square both sides. If I square the modulus what would be the answer like z^2 what? Once I get past this I am confident I can solve the question.

Sketch the circle described by the equation

|z − u| = 2
Determine the points of intersection (if any) of this circle and the two axes, real and imaginary.

where u = −1 + j√3
Use the rule that the square of the modulus is the sum of the squares of the real and imaginary parts - it is the length of the hypotenuse of a right triangle.

z = x + jy
u = -1 + j√3
z - u = (x + 1) + j(y - √3)
|z - u|^2 = . . . = 4
 
Hey so I have another complex numbers question. My main issue is I am getting confused with the modulus sign. In this case how would I get rid of it? Square both sides. If I square the modulus what would be the answer like z^2 what? Once I get past this I am confident I can solve the question.

Sketch the circle described by the equation

|z − u| = 2
Determine the points of intersection (if any) of this circle and the two axes, real and imaginary.

where u = −1 + j√3

Would you know the answer to the question (when u =0):

Sketch the circle described by the equation

|z| = 2

Determine the points of intersection (if any) of this circle and the two axes, real and imaginary.
 
Well when u = 0 it is going to be a circle which starts on the x-axis with a radius of 4

So this is what I have done. Am I on the right track?

|z+1 -j√3| = 2

|(z+1)^2
(-j√3)^2|= 2^2

z^2 + 2z +1 -3j = 4

z(z+2) = 3-3j

z= 3-3j and z= 1-3j
 
Last edited:
Can you draw the circle - where is its center?

Where does it intersect the axes?
 
Well when u = 0 it is going to be a circle which starts on the x-axis with a radius of 4

So this is what I have done. Am I on the right track?

|z+1 -j√3| = 2

|(z+1)^2
(-j√3)^2|= 2^2

z^2 + 2z +1
-3j = 4 ............... how did you get that?

z(z+2) = 3-3j

z= 3-3j and z= 1-3j
.
 
Well when u = 0 it is going to be a circle which starts on the x-axis with a radius of 4

What does that mean?

Also the radius is not 4 - it would be 2

The circle has its center at the origin.

So this is what I have done. Am I on the right track?

|z+1 -j√3| = 2

|(z+1)^2
(-j√3)^2|= 2^2

z^2 + 2z +1 -3j = 4

z(z+2) = 3-3j

z= 3-3j and z= 1-3j

.
 
Sorry I just read Dr.Phils comment and I realized everything I had done was wrong.

What I meant about the circle starting on the axis is, if for instance it was x^2 = j, it starts at pi/2 or the y axis
 
So the centre of the circle is at (-1,√3) with a radius of 2.

basically this equation says it:
(x+1)^2 +j(y-
√3)^2 = 4
 
So the centre of the circle is at (-1,√3) with a radius of 2.

basically this equation says it:
(x+1)^2 +j(y-
√3)^2 = 4 ← This is incorrect

The equation of the circle is (no j):


(x+1)^2 + (y-√3)^2 = 4

So where does it intersect the x-axis and the y-axis?
 
My mind went blank I don't remember how to find where it intersects the axis. Clearly I can't make x or y =0.

You tell me please
 
My mind went blank I don't remember how to find where it intersects the axis. Clearly I can't make x or y =0.

You tell me please

Yes - you can. That is how you find intercepts with axes.
 
My mind went blank I don't remember how to find where it intersects the axis. Clearly I can't make x or y =0.

You tell me please
That is not clear to me! In fact that is exactly what you do.

The equation for y-axis is x=0\displaystyle x = 0. If the circle does intersect the y-axis, then x will be 0 at that point. So set x=0 and determine whether any value of y satisfies the equation of the circle.

Do you have the equation of the circle as Subhotosh Khan wrote it for you? Note that it is expressed purely in Real numbers on the xy-plane.
 
(1)^2 + (y-√3)^2 = 4

(y-√3)^2 = 3

y^2 -2
√3y +3 = 3

y(y -
2√3) = 0

y =0 and y =
2√3

so at these 2 points it intersects with the y axis
 
(1)^2 + (y-√3)^2 = 4

(y-√3)^2 = 3

y^2 -2
√3y +3 = 3

y(y -
2√3) = 0

y =0 and y =
2√3

so at these 2 points -
(0,0) & (0,2√3) - it intersects with the y axis

Now find x-intercepts
 
(1)^2 + (y-√3)^2 = 4

(y-√3)^2 = 3

y^2 -2
√3y +3 = 3

That's the "hard way". Just take the square root of both sides: from (y3)2=3\displaystyle (y- \sqrt{3})^2= 3 we can immediately get y3=±3\displaystyle y- \sqrt{3}= \pm\sqrt{3}. If y3=3\displaystyle y- \sqrt{3}= \sqrt{3} then y=23\displaystyle y= 2\sqrt{3}. If y3=3\displaystyle y- \sqrt{3}= -\sqrt{3} then y=0\displaystyle y= 0

2√3) = 0

y =0 and y =
2√3

so at these 2 points it intersects with the y axis
 
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