Easiest to prove using the trigonometric/exponential form of complex numbers. (1-i)/(1+i)=√2exp(-iπ/4)/√2exp(iπ/4)=exp(-iπ/2) so log[(1-i)/(1+i)]=-iπ/2 and hence the result. (Of course, it's all modulo 2πi, because exp(x+2πi)=exp(x).)
-
-
-
You can also use the 'multiply by the conjugate' trick to see what the fraction is: (1-i)/(1+i)=[(1-i)/(1+i)][(1-i)/(1-i)]=(1-i)²/2=[1-2i+(-i)²]=-i. Then write -i as exp(-iπ/2), take the log and you're done.
- 4 more replies
New conversation -
-
-
I've never seen that tanh-1 => tan-1 manipulation, how does that happen?
- 2 more replies
New conversation -
-
-
I believe the multiplication in the right should be (1-i)/(1+I) * (1-i)/(1-i)
- 1 more reply
New conversation -
-
-
The irrational, the imaginary, the real and the transcendental put all together...
#lovelyThanks. Twitter will use this to make your timeline better. UndoUndo
-
-
-
I see a lot of demonstrations here. But what i'd like to know is how he, Fagnano, demonstrated that with the math tools of his time. This should be interesting !
Thanks. Twitter will use this to make your timeline better. UndoUndo
-
-
-
What is the base of this log?
- End of conversation
New conversation -
Loading seems to be taking a while.
Twitter may be over capacity or experiencing a momentary hiccup. Try again or visit Twitter Status for more information.