Question

Prove that any complex number with modulus 1 (exept -1) can be represented in the form: (1+it)/(1-it) where t is a real

Answer 1

both numerator and denominator has the same moduli
(1+it)/(1-it) ... so this term has modulus 1

Answer 2

\[\frac{1+it}{1-it} \frac{1+it}{1+it}=\frac{1-t^2+2ti}{1+t^2}=\frac{1-t^2}{1+t^2}+\frac{2t}{1+t^2} i\]

Answer 3

and?

Answer 4

or any z = a + ib such that a^2 + b^2 = 1

|1 - it| = |1 + it| = sqrt(1 + t^2)
arg(1 - it) = - arg(1 + it)

e^(2arg(1+it)) = sqrt(a^2 + b^2)e^(2*(1/2)arg(z))) = sqrt(a^2+b^2)e^(1/2*arg(z))/(sqrt(a^2+b^2)e^(-1/2 arg(z)))
\
\[ \frac{\sqrt{a^2 + b^2}e^{{1\over 2}arg(z)}}{\sqrt{a^2 + b^2}e^{{-1\over 2}arg(z)}} = \frac{\sqrt{1 + (b/a)^2}e^{{1\over 2}arg(z)}}{\sqrt{1 + (-b/a)^2}e^{{-1\over 2}arg(z)}} \]

Answer 5

still that 1/2 is troubling.

Answer 6

nice

Answer 7

tx @experimentX

Answer 8

Oo

Answer 9

hold on ...
\[ \huge a + ib = e^{i\arctan(b/a)} =\\
\huge \frac{1 + i {\sin (1/2 \arctan(b/a))\over \cos (1/2 \arctan(b/a))}}{ 1 - i {\sin (1/2 \arctan(b/a))\over \cos (1/2 \arctan(b/a))}} \]

Answer 10

\[\frac {\sqrt{1+t ^{2}}e ^{\frac{1}{2}argz}}{\sqrt{1+t ^{2}}e ^{-\frac{1}{2}argz}}=e ^{\arg(z)}=z\]

Answer 11

thx again

Answer 12

For -1, if you use the above formula, you get
-1 = 1

Answer 13

no probs.