complex number question (1 Viewer)

[Ragerunner]

Multiplying a complex number z by e^i*theta rotaties it anticlockwise about the origin throjugh an angle theta.

Find in (a+ib form) the complex number obtained by rotating 1- squareroot of 3 i anticlockwise about the origin through an angle pi/3..

also

Find all complex square roots of 3 - 4i by solving (x + yi)^2 = 3 - 4i, for x,y real.

Thanks.

 

[Ragerunner]

First one i got (1 + squareroot 3 i) - Is that correct?

I don't know how to do the second one..

 

[George W. Bush]

x^2 - y^2 + 2xyi = 3 - 4i
Equate real and imaginary parts then solve using simultaneous equations

 

[Ragerunner]

hmm ok thanks.

How abouth this one? It's similar to the first one i posted but im not sure whether im correct or not.

Multiplying a complex number z by e^i*theta rotaties it anticlockwise about the origin throjugh an angle theta.

Find in (a+ib form) the complex number obtained by rotating 1+ squareroot of 3 i anticlockwise about the origin through an angle pi/6..

 

[George W. Bush]

e^ix = cisx - I don't know why whereever you got the question from is referring to cisx as e^ix, as e^ix isn't in the 4u syllabus, but nevertheless.

1 + sqrt3 = 2cis pi/3
Now recall that cis(A) * cis (B) = cis(A+B)
now, 2cis(pi/3) * cis (pi/6) = 2cis(pi/2)
You need to convert that back to a+ib form
cos pi/2 = 0, sin pi/2 = 1
So 2cis(pi/2) = 2i

 

[Constip8edSkunk]

2e^(i*pi/3) * e^(i*pi/6)
=2e^(i*pi/2)
=2i

edit: beat me to it... yeah this is for uni, good luck on the algebra test 2moro

 

[Ragerunner]

Thanks for the help!

Good Luck on your test too.