Finished your HSC? Check out our University Discussion, Non-School forums or help out next year's HSC students!
YOU can help the next generation of students in the community! Share your trial papers and notes on our Notes & Resources page

De Moivre's Theorem (1 Viewer)

Thread starter ashokkumar
Start date Jan 24, 2010

A

ashokkumar

Mr

#1

Express the following in x+ yi form:

(3 - 4i)^5 / (4 + 3i)^3

Azamakumar

bannèd

#2

mind
fuck

SeCKSiiMiNh

i'm a fireball in bed

#3

dont you just change it to mod/arg form, divide it, then change it back???

gella

Member

#4

OMG ashleeey

gurmies

Drover

#5

$\text{Note that} \,\, 4+3i=i\left ( 3-4i \right ) \\\\ \therefore \frac{\left (3-4i \right )^{5}}{\left (4+3i \right )^{3}}=\frac{\left (3-4i \right )^{5}}{\left (3-4i \right )^{3}\left ( i \right )^{3}} \\\\ = -\frac{\left ( 3-4i \right )^{2}}{i} = -\frac{9-24i-16}{i} \\\\ = \frac{7+24i}{i} = 24-7i$

A

ashokkumar

Mr

#6

wow that was efficient. thanks

MongMan

Member

#7

It's a loaded question.

However you should still know how to solve this using De Moivres.

Last edited: Jan 26, 2010

MongMan

Member

#8

(mod1)^5/(mod2)^3 * cis(5alpha - 3beta) easy peasy with de Moivre

Last edited: Jan 26, 2010

You must log in or register to reply here.

Share:

Facebook Twitter Reddit Pinterest Tumblr WhatsApp Email Link

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

Top