Complex numbers loci problem (Find square roots of 2 + i*sqrt{5}; hence...)

[annaanna]

Need help for last part, first of all I dont understand how 2 roots
can now be described as one single thing ( is alpha a line? i dont understand
how that makes sense either), and then it wants the perpendicular bisector but
from the description of alpha as a line isnt it at basically (0,0), but
isnt z assumed to be that? I dont really understand.

 

[stapel]

Are they saying that the two roots have become one, or are they saying that they're working with the root that is on the specified interval?

 

[Dr.Peterson]

( is alpha a line? i dont understand
how that makes sense either)

For part iii, you should have four points, one in each quadrant.

Alpha is a complex number (the root of the equation that is in the first quadrant), not a line. In talking about a line perpendicular to alpha, they are just roughly describing the appearance of the correct graph, apparently meaning the perpendicular bisector of the segment from the origin to alpha. You should not be trying to reverse-engineer your solution from what they say here.

Do you see why the locus of [imath]|z-\alpha|=|z|[/imath] would be that perpendicular bisector?

then it wants the perpendicular bisector but
from the description of alpha as a line isnt it at basically (0,0), but
isnt z assumed to be that?

What are you saying here? What is at (0,0)? I don't understand.

 

[annaanna]

Are they saying that the two roots have become one, or are they saying that they're working with the root that is on the specified interval?

First option I wouldn't understand and second option there are no roots on whatever alpha is meant to be

 

[annaanna]

For part iii, you should have four points, one in each quadrant.

oh yeah I forgot about the complex conjugates

Alpha is a complex number (the root of the equation that is in the first quadrant), not a line. In talking about a line perpendicular to alpha, they are just roughly describing the appearance of the correct graph, apparently meaning the perpendicular bisector of the segment from the origin to alpha. You should not be trying to reverse-engineer your solution from what they say here.

Omg i had to read this a couple of times but i see what they mean now thank you so much.

Do you see why the locus of [imath]|z-\alpha|=|z|[/imath] would be that perpendicular bisector?

What are you saying here? What is at (0,0)? I don't understand.

I didnt understand the whole root part so i was going off my diagram which was this ...I assumed alpha basically made a line at 0,0 so i didnt understand what i was even meant to subtract from z

 

[Dr.Peterson]

I didnt understand the whole root part so i was going off my diagram which was this ...I assumed alpha basically made a line at 0,0 so i didnt understand what i was even meant to subtract from z

Alpha is the root in the first quadrant, namely [imath]\frac{\sqrt{10}}{2}+\frac{\sqrt{2}}{2}i[/imath].

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The locus of [imath]|z-\alpha|=|z|[/imath] is the set of all points z that are the same distance from [imath]\alpha[/imath] as from the origin. Do you see that?

What is the set of points equidistant from two given points?