Question

Please simplify:)
Its answer is [(2ab)/(a^2-b^2)].

Answer 1

here i=iota, a complex number of form (0,1).

Answer 2

\[\tan\left[i\log\left(\frac{a-ib}{a+ib}\right)\right]\]

Answer 3

Pl start with denoting a + ib in the form r( cos x + i sin x)

Then equate real and imaginary part. Try.

Answer 4

@UnkleRhaukus plz help:) & ya that is the question.

Answer 5

i have not studied complex analysis, and i have not done a problem like this before,

Answer 6

k! np:)

Answer 7

ya it is the answer I know; but I haven't solved it:(

Answer 8

@Mertsj plz help:)

Answer 9

start with
\[\log(a-bi)-\log(a+bi)\] then by definition \(\log(z)=\ln(z)+i\theta\) so you get
\[\ln(\sqrt{a^2+b^2}+i\theta -(\ln(\sqrt{a^2+b^2})-i\theta)\] the last part because if arg \(a+bi=\theta\) then arg\(a-bi=-\theta\)
this adds to \(2i\theta\) multiply by \(i\) get \(-2\theta\) and you want the tangent \[\tan(-2\theta=-\tan(2\theta)\] and \[\tan(\theta)=\frac{b}{a}\]
so now double angle formula should give you the result

i might have made a mistake in calculation, so check it, but this i am fairly sure that this is the right method

Answer 10

Put a=r cosA, b==r sinA
Expression= tan [i. log (cosA-i sinA)/cosA+i sinA)]
=tan [i. log (e^-iA)/(e^iA)]
=tan[i. log e^(-2iA)]
=tan[i.(-2iA)]
=tan 2A
=2 tanA/(1-tan^2 A)
=2.(b/a) / [1-(b/a)^2]
=2ab/a^2 +b^2

Answer 11

k! thanx a lot:)

Answer 12

UR welcome.