# Extra_3 - Extra Problems Sheet 3 Stephen Taylor May 4 2005 1 Define f C ∪{∞ → C ∪{∞ by f z = az b cz d where a b c d are fixed complex

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Unformatted text preview: Extra Problems Sheet 3 Stephen Taylor May 4, 2005 1. Define f : C ∪ {∞} → C ∪ {∞} by f ( z ) = az + b cz + d where a, b, c, d are fixed complex numbers. Show that f is one-to-one and onto if ad- bc 6 = 0 . What happens if ad- bc = 0? One- to- one : Suppose f ( z ) = f ( w ) where w 6 = z . Then az + b cz + d = aw + b cw + d ⇒ ( az + b )( cw + d ) = ( aw + b )( cz + d ) ⇒ azd- adw = bcz- bcw ⇒ ad ( z- w ) = bc ( z- w ) ⇒ ( z- w )( ad- bc ) = 0 So if ad- bd 6 = 0 then z = w , which shows that f is one-to-one. Consider the case f ( z ) = ∞ f ( w ). Then cz + d = 0 = cw + d . c 6 = 0, otherwise ad- bc = 0. So z = w , and we have shown then function is one-to-one. Onto : Let w = f ( z ) ∈ C ∪ {∞} . Then w = az + b cz + d ⇒ wcz + wd = az + b ⇒ z ( wc- a ) = b- wd ⇒ z = b- wd wc- a So the above shows that f is onto everywhere except in the case where w = a/c . Consider the case f ( z ) = a/c . This happens iff z = ∞ which shows the function is onto everywhere. 2. Show that if z, w ∈ C ∪ {∞} , then d ( z, w ) = d 1 z , 1 w We note the metric on C ∪ {∞} is defined to be 1 d ( z, w ) = 2 | z- w | p ( | z | 2 + 1)( | w | 2 + 1) So by definition d 1 z , 1 w = 2 | z- 1- w- 1 | p ( | z- 1 | 2 + 1)( | w- 1 | 2 + 1) = 2 | zw || z- 1- w- 1 |...
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## This note was uploaded on 10/19/2009 for the course MATH 814 taught by Professor Cong during the Three '09 term at University of Adelaide.

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Extra_3 - Extra Problems Sheet 3 Stephen Taylor May 4 2005 1 Define f C ∪{∞ → C ∪{∞ by f z = az b cz d where a b c d are fixed complex

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