MedVision ad

Complex numbers q (1 Viewer)

HazzRat

H̊ͯaͤz͠z̬̼iẻͩ̊͏̖͈̪
Joined
Aug 29, 2021
Messages
1,249
Gender
Male
HSC
2024
juytrhrgefwd.PNG

In this question, you have to prove |z| = 1 given z is a complex number satisfying |2z - 1| = |z - 2|

Worked solution is above but I have two questions:
1) why does squaring both sides of the equation work to get rid of the modulus sign
2) how did they get from step 5 to step 6 (the one's I circled in red)

Ty in advance
 

TankKuno

Well-Known Member
Joined
Nov 5, 2022
Messages
286
Gender
Male
HSC
2023
View attachment 41023

In this question, you have to prove |z| = 1 given z is a complex number satisfying |2z - 1| = |z - 2|

Worked solution is above but I have two questions:
1) why does squaring both sides of the equation work to get rid of the modulus sign
2) how did they get from step 5 to step 6 (the one's I circled in red)

Ty in advance
because the modulus means the root of the additoon of the real and imaginary components squared, so squaring both sides gets rid of the surd
 

HazzRat

H̊ͯaͤz͠z̬̼iẻͩ̊͏̖͈̪
Joined
Aug 29, 2021
Messages
1,249
Gender
Male
HSC
2024
It’s just the modulus of a complex
How do u find it? Sqrt(re z sqr+ Imz sqr)
Then if u square that it just gets rid of the root
because the modulus means the root of the additoon of the real and imaginary components squared, so squaring both sides gets rid of the surd
Ty ty - didn't know a modulus is a surd
 

carrotsss

New Member
Joined
May 7, 2022
Messages
4,435
Gender
Male
HSC
2023
Someone teach me latex pls
some parts are outdated but:
 

HazzRat

H̊ͯaͤz͠z̬̼iẻͩ̊͏̖͈̪
Joined
Aug 29, 2021
Messages
1,249
Gender
Male
HSC
2024
some parts are outdated but:
 

HazzRat

H̊ͯaͤz͠z̬̼iẻͩ̊͏̖͈̪
Joined
Aug 29, 2021
Messages
1,249
Gender
Male
HSC
2024
Another one:
How would you graph this:
hrtegrwfw.PNG
 

Luukas.2

Well-Known Member
Joined
Sep 21, 2023
Messages
444
Gender
Male
HSC
2023
Another one:
How would you graph this:
View attachment 41060
You can use to show it rearranges to , and so is the region .

However, thinking of the expression as a vector, the interpretation of the statement is:

is positioned such that the distance from to is less than 2 units​

Since is located at the reflection of in the real axis, is a purely imaginary vector with its midpoint on the real axis.

So, if the length of this vector must be less than 2 units, then must lie within 1 unit (vertically) from the real axis.

Thus, the region covers all possible solutions for .

---

These kinds of vector interpretations can make more algebraically complicated problems easier.

For example: Find the maximum value of given that and .

The answer is , arising from the four values and that correspond to the four corners of the square that defines the region within which must lie.
 

HazzRat

H̊ͯaͤz͠z̬̼iẻͩ̊͏̖͈̪
Joined
Aug 29, 2021
Messages
1,249
Gender
Male
HSC
2024
Another one:
I’m just working ahead atm and idk how to do these sorts of questions. For example, how would you do 9(c)?
1698644069795.png
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top