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 ] =
[y * y ] =
[exp (-2πi(Kx )] [exp (2πi(Kx )] = exp(Kx) exp(Kx)
= exp(Kx + Kx) = exp (2Kx)
= 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 ) ?)
r exp(i q ) ?)
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