## Thursday, March 7, 2013

### Solutions to Complex Conjugate Problems

a) (5 + i)(4 - 3i)(1 - i)

Take: (5 + i)(4 - 3i) = 20 +4i -15i -3(i)(i) = -11i +20 +3 = 23 -11i

Then: (23 – 11i)(1 –i) = 23 -11i -23i -11 = 12 – 34i

So complex conjugate = 12 + 34i

b) (1 + i) / (1 - i)

We have, since arctan(y/x) = arctan(1) and q = 45o

(1 + i) = Ö2 [cos (45) + isin(45)] =  Ö2 cis(p/4)

So, z1 = Ö2 exp (ip/4 )

The reader should be able to easily show that (1- i) is the same except for the argument, which must be (- q) since we have arctan (-1/1).

Thus:  z2 = (1 – i) = Ö2 exp (i(-p/4) )

Then: dividing:

(z1/z2) = (Ö2 cis(p/4)/ Ö2cis(-p/4)) =   cis{p/2}

= cos (p/2) + i sin(p/2) = i

So the complex conjugate is: -i

c)
(1 – i)3

(1 – i)3   = (1- i)(1- i)(1- i)

(1 – i)(1 – i) = 1 –2i +(-1) = 1-1 -2i = -2i

Finally: (1 – i) ((-2i) = -2i +2(-1) = -2i -2 or -2 – 2i

So the complex conjugate = -2 +2i

2) A quantum wave function is expressed:  y = exp [2πi(Kx )]

Obtain the complex conjugate (y*) and hence or otherwise find the probability amplitude  [y  * y  :

y * = exp [- 2πi(Kx )]

[y  * y  ]  =

[exp (-2πi(Kx )] [exp (2πi(Kx )] =  exp(Kx) exp(Kx)

= exp(Kx + Kx) = exp (2Kx)

Note: exp [2πi] = exp [-2πi] = 1

(Can you verify the above using: cos(q )  + isin(q ) =

r exp(i q ) ?)