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April 21, 2021, 05:43:11 pm

### AuthorTopic: Help with Complex number Loci questions  (Read 553 times) Tweet Share

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#### nat_21

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##### Help with Complex number Loci questions
« on: March 05, 2021, 10:24:31 pm »
0
Hi all,
I tried to expand the algebra but it does not work, maybe I went wrong somewhere. Could anyone show me how to solve this question? (IMG_3199)
For example 3.13, how do you find the radius? (IMG_3200)
Thank you very much.

#### fun_jirachi

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##### Re: Help with Complex number Loci questions
« Reply #1 on: March 05, 2021, 11:55:31 pm »
+1
Hey

For the first question, if you're sure you went wrong with your algebra but that your method is correct, it might be easier for you to post your working and for us to have a look.

For the second question, the following two circle geometry theorems might be helpful to consider:
1. The angle subtended by two points on the circumference to another point on the circumference is half that of the same two points at the centre of the circle.
2. A line from the centre of the circle will bisect any chord at a right angle.

Using these with a little algebra should allow you to find the centre reasonably easily.

Hope this helps
Spoiler
HSC 2018: Mod Hist [88] | 2U Maths [98]
HSC 2019: Physics [92] | Chemistry [93] | English Adv [87] | 3U Maths [98] | 4U Maths [97]
ATAR: 99.05

UCAT: 3310 - VR [740] | DM [890] | QR [880] | AR [800]
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#### nat_21

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##### Re: Help with Complex number Loci questions
« Reply #2 on: March 06, 2021, 03:17:49 pm »
0
Hey

For the first question, if you're sure you went wrong with your algebra but that your method is correct, it might be easier for you to post your working and for us to have a look.

For the second question, the following two circle geometry theorems might be helpful to consider:
1. The angle subtended by two points on the circumference to another point on the circumference is half that of the same two points at the centre of the circle.
2. A line from the centre of the circle will bisect any chord at a right angle.

Using these with a little algebra should allow you to find the centre reasonably easily.

Hope this helps
Hi, thank you very much, I understand the second question now. For the first question, I am not too sure did I approach it in the right way.  But here is my working, I didn't finish it because I don't know what to do with it. I really appreciate your help.

#### fun_jirachi

• MOTM: AUG 18
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##### Re: Help with Complex number Loci questions
« Reply #3 on: March 06, 2021, 04:16:26 pm »
+1
I think from the outset your method is wrong - you're assuming that $w$ is purely imaginary then finding the locus some $z$ such that $z=x+iy$. It should be done in the opposite way ie. you choose some $z$ such that $z=x+iy$ and hence determine the conditions on $x, y$ (ie. the locus!) such that $w$ is imaginary.

Realising the denominator immediately doesn't help too much - we ideally want to express w in terms of $z$ or $x, y$ where $z = x+iy$.
Rearranging a bit will get you that $w=\frac{1}{3} \frac{z+4i}{(z-1)} = \frac{1}{3} \frac{x+i(y+4)}{(x-1)+iy}$ - realising the denominator here will give you some expression for $w$ such that you can split it into its real and imaginary parts. After doing this, it should be trivial enough to obtain an expression for $w$ such that $w$ is purely imaginary ie. $\text{Re}(w) = 0$.

Hope this helps
Spoiler
HSC 2018: Mod Hist [88] | 2U Maths [98]
HSC 2019: Physics [92] | Chemistry [93] | English Adv [87] | 3U Maths [98] | 4U Maths [97]
ATAR: 99.05

UCAT: 3310 - VR [740] | DM [890] | QR [880] | AR [800]
Subject Acceleration (2018)
UCAT Question Compilation/FAQ (2020)

#### nat_21

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• Posts: 3
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##### Re: Help with Complex number Loci questions
« Reply #4 on: March 06, 2021, 05:05:31 pm »
+1
I think from the outset your method is wrong - you're assuming that $w$ is purely imaginary then finding the locus some $z$ such that $z=x+iy$. It should be done in the opposite way ie. you choose some $z$ such that $z=x+iy$ and hence determine the conditions on $x, y$ (ie. the locus!) such that $w$ is imaginary.

Realising the denominator immediately doesn't help too much - we ideally want to express w in terms of $z$ or $x, y$ where $z = x+iy$.
Rearranging a bit will get you that $w=\frac{1}{3} \frac{z+4i}{(z-1)} = \frac{1}{3} \frac{x+i(y+4)}{(x-1)+iy}$ - realising the denominator here will give you some expression for $w$ such that you can split it into its real and imaginary parts. After doing this, it should be trivial enough to obtain an expression for $w$ such that $w$ is purely imaginary ie. $\text{Re}(w) = 0$.

Hope this helps
It works! I found the answer, thank you very much!!