Complex Numbers Question (1 Viewer)

[HSCstudent08]

QUESTION: Find the complex number z with the least positive argument satisfying the condition | z - 5i | ≤ 3 (include a diagram)

thanks to anyone who can answer this question

 

[Trebla]

Equation of region is x² + (y - 5)² ≤ 9

Draw a tangent to the circle from the origin which gives the lowest argument. You could try sub y = mx into x² + (y - 5)² = 9 and find values of m for which the discriminant is zero. There should be two values, choose the one with the positive value (positive gradient for smallest argument). Once you've found m, solve simultaneously to find point of intersection, hence the complex number.

There's probably a quicker way to do it (without referring to an accurate diagram) though I can't think of it from the top of my head at the moment...

 

[Aerath]

Equation of region is x² + (y - 5)² ≤ 9

Draw a tangent to the circle from the origin which gives the lowest argument. You could try sub y = mx into x² + (y - 5)² = 9 and find values of m for which the discriminant is zero. There should be two values, choose the one with the positive value (positive gradient for smallest argument). Once you've found m, solve simultaneously to find point of intersection, hence the complex number.

That's probably how I'd do it - I can't think of any other way. =\

 

[lolokay]

Equation of region is x² + (y - 5)² ≤ 9

Draw a tangent to the circle from the origin which gives the lowest argument.

that should be all you need. just draw in the radius of the circle to that point and you have a right angled triangle, so you can get the angle

 

[HSCstudent08]

This is the diagram i Had:

when you're working out least positibe argument, isn't that just the angle? use tan or something to get your angle?

can anyone else find a quicker method? ans = 12/5 + 16/5 i

 

[Pwnage101]

yeh i like to use the circle geometry way, with the fact that a radius and tangent at a point make a right angle, and you know 2 lengths of teh triangle (the radius and teh distance from O to the centre), so then use trig to find the angle, and the argument is pi/2 minus that, in this case.

the y=mx method is also good i found.

Either way works =)

 

[tommykins]

The questions asks for the complex number z, not it's argument.

To find the points you have a right angled triangle with 4 as the hypotenuse (refer to above diagram).

Since you have the argument @ (assuming so), simply for the x co-ordinate you have cos@ = x/4
x = 4cos@

Same as y, y = 4sin@

 

[Aerath]

Is it wrong to just put z = 4cis@?

 

[tommykins]

depends if it asks for cartesian or not.

haha been a while since i've seen cis@, so....eurgh.

 

[gurmies]

In theory 4cis@ is A SINGLE complex number so it would be strange to get penalised for that. I would put 4cis@ or/ x + iy to cover both, it doesn't take much time...