Complex Numbers-HELP PLEASE!!!! (1 Viewer)

[chris_power_96]

(i) Find the equation of the locus of z given |z-2i|=Im(z)
(ii) Find the maximum and minimum values of argz
(iii) Find z in the form a+ib when argz takes its minimum value

I could do (i) and my answer was y=(x^2)/2 +2, but I couldn't complete ii or iii

Any help would be appreciated

 

[chris_power_96]

For part i this is what I did:
let z=x+iy
New Equation: |x+i(y-2)|=y
(x^2 +(y-2)^2 )^(1/2)=y
x^2 + (y-2)^2 =y^2
x^2 + y^2 -4y +4=y^2
4y=x^2 + 4
y= (x^2)/4 +1


I know this is different from my initial equation (I found where I made a mistake), but could you explain what is wrong with that working out

 

[Sy123]

For part i this is what I did:
let z=x+iy
New Equation: |x+i(y-2)|=y
(x^2 +(y-2)^2 )^(1/2)=y
x^2 + (y-2)^2 =y^2
x^2 + y^2 -4y +4=y^2
4y=x^2 + 4
y= (x^2)/4 +1


I know this is different from my initial equation (I found where I made a mistake), but could you explain what is wrong with that working out

I made a crucial mistake somewhere, I apologize -> rewriting my solution now. (and for part i you are right)

 

[Sy123]

Ok since latex doesnt work. I will walk you through the steps.

Part i) Yes you are correct

y= (x^2)/4 + 1

ii)

Now we are asked to find a minimum argument, this can be observed that the minimum argument is the point where it is a tangent to the parabola from the origin.
So our objective is to find the equation of the line that intersects our parabola ONLY once.
The only way this is the case is tangent.

So our first equation is the equation of the parabola.

y=(x^2)/4 + 1

What is the equation of the line from the origin to the point on the parabola with min. argument?

y = mx (for some gradient m, we need to actually find m)

Let us equate our equations:

mx = x^2/4 + 1

x^2 - 4mx + 4 = 0

Now, this equation ONLY has one solution. So lets equate discriminant to zero.

16m^2 - 16 =0

m= 1, m= -1

One gives us the gradient for the MAXIMUM argument
One gives us gradient for MINIMUM argument.

So the angle the line y= x makes with the x-axis is 45 degrees, hence 45 degrees is our minimum argument
Similarly for y= -x, the maximum argument is 135 degrees.

iii)

Now we have the minimum argument, we must also find the minimum modulus to find our answer in terms of a+ib

So we have the equations:

y=x^2/4 + 1
y=x

Equate again.

x^2/4 -x + 1 =0
x^2 - 4x + 4 =0

x=2
y=2

That is the point of intersection, so the modulus, is the distance from (0,0) to (2,2) which can be found from pythagoras.
It ends up being 2root2

So, we have found modulus 2sqrt2 argument pi/4

Sub this in

z = 2sqrt 2 ( cos 45 + i sin 45 )

z= 2sqrt 2 ( 1/sqrt 2 + i 1/sqrt 2)

z= 2+2i

 

[Carrotsticks]

Part (ii) [Alternative to Sy123's geometric solution]