• Congratulations to the Class of 2024 on your results!
    Let us know how you went here
    Got a question about your uni preferences? Ask us here

Complex numbers help (1 Viewer)

Blackmancan

New Member
Joined
Oct 6, 2011
Messages
25
Gender
Undisclosed
HSC
2013
I just started complex numbers and got stuck on these:
z1=2-3i , z2=1+4i

q1. 1/z2

q2. (z1)^3 - (z2)^3

:/
 
Last edited:

TheCardician

Member
Joined
Nov 4, 2009
Messages
81
Location
Menai, Sydney, NSW
Gender
Male
HSC
2011
Q1) You need to multiply the numerator and denominator by the conjugate of 1-4i which is 1+4i; so:1+4i/(1+4i)(1-4i) = 1+4i/1+16 = (1+4i)/17
Q2) You can either express it as a sum of two cubes, or just expand them both as is: (2-3i)^3 = 8-36i+54i^2 -27i^3 = 8-36i-54+27i = 62-63i
(1+4i)^3 = 1+12i+48i^2 +64i^3 = 1+12i-48-64i = -47-52i

Now, 62-63i - 47-52i = 15-115i

Hope this helps.

EDIT: Realized in Q1 I used z1 instead of z2 -_-
 
Last edited:

kooliskool

Member
Joined
Nov 23, 2006
Messages
138
Gender
Male
HSC
2007
Hi, so for q1:



First you should think of to get the real part and the imaginary part out, you need to cancel the i in the denominator somehow, which is done similar to the method in 2U in rationalising the denominator:





Since




lols, for q2, there you go, some guy got to it before me :). Although it's difference of two cubes, not some of two cubes..... You better do it by what he did though, it's faster to just expand it, since you need to ultimately expand it anyway.
 
Last edited:

D94

New Member
Joined
Oct 5, 2011
Messages
4,423
Gender
Male
HSC
N/A
So you just expanded both z1 and z1 the 3U style? What if it were something like z1^5-z2^5?
You could do it a few ways; expand using binomial, or more conveniently, change it to mod-arg form and apply the de Moivre's theorem, and then change it all back to x+iy form.
 

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

Top