ce 03 amaths

First of all, simplify the complex number (1 + 3i)/(1 - 2i)

(1 + 3i)/(1 - 2i)

= (1 + 3i)/(1 - 2i) X (1 + 2i)/(1 + 2i)

= (1 + 3i)(1 + 2i) / [12 - (2i)2]

= [1 + 5i + 6i2] / (1 - 4i2)

= (1 + 5i - 6) / (1 + 4)

= -1 + i

So, [(1 + 3i)/(1 - 2i)]20

= (-1 + i)20

= (√2)20 (-1/√2 + 1/√2 i)20

= 210 [cos(3pi/4) + isin(3pi/4)]20

By DeMoivre's theorem,

[cos(3pi/4) + isin(3pi/4)]20

= [cos(20)(3pi/4) + isin(20)(3pi/4)]

= cos(15pi) + isin(15pi)

= -1

So, [(1 + 3i)/(1 - 2i)]20

= 210 (-1)

= -1024


2009-04-01 11:04:40 補充：
This is out of the current amaths syllabus.