Question

If a+ib=tanh(x+i π/4),prove that a^2+b^2=1

Answer 1

what is the shortest way....taking log on both sides? didn't help much...

Answer 2

first expand the right side to the definition of tanh

Answer 3

i would do it the difficult way :P

Answer 4

\(\Large \dfrac{e^y-e^{-y}}{e^y+e^{-y}}\)

where y =x+i π/4
then ?

Answer 5

\(\Large e^y = e^{x+i\pi/4} = e^x e^{i\pi/4} = ...\)

Answer 6

well i would probably condense it to -i/cot(i*y)

Answer 7

oh and then ?

Answer 8

any formula for cot iy ?

Answer 9

so we have
a + bi = -i/cot(i*y)
multiply both sides by i
ai - b = 1/cot(i*y)
ai - b = tan(i*y)
ai - b = tan(ix - pi/4)

Answer 10

what do you think you can do from there?

Answer 11

tan (a-b) formula ?
but that will make the denominator have the complex term...

Answer 12

yeah, and there's an i on the left side of the equation as well. we can multiply things out to make it work (i think, i may have done it wrong but i think it'll work :P trust)

Answer 13

sure! lets try


ai-b = (tan ix-1)/(1-tanix) = -1

whoa...is that correct ?

Answer 14

yes! (tan (ix)-1)/(1-tan(ix)) = -1

Answer 15

thanks!

Answer 16

no worries. you can prove it with the given info right?

Answer 17

ofcourse :)

Answer 18

the first step was important!

Answer 19

yeah, i hate working with tanh identities. much easier to just expand tanh because -i/cot(ix) is a very pretty way of putting tanh