Question

Plz help with this: -

Answer 1

\[\tan [i \log_e{\frac{a-ib}{a+ib}}]\]

Answer 2

Simplify: -

Answer 3

Answer is : -
\[\frac{2ab}{a^2-b^2}.\]

Answer 4

It is a complex number problem.

Answer 5

@TuringTest @Zarkon @myininaya Plz help

Answer 6

you can write any complex number in the form:
\[z=x+iy=r e^{i \theta}\]

Answer 7

you got \[a-ib/a+ib=(a-ib)(a+ib)/a^2+b^2\]

if you multiply and divide by conjugate

Answer 8

\[\frac{a-ib}{a+ib}=\frac{(a-ib)(a-ib)}{(a+ib)(a-ib)}=\frac{a^2-b^2-2abi}{a^2+b^2}\]

Answer 9

k:)
I will do it nw:)

Answer 10

and you gey doing the above product the expresion mukushla said

Answer 11

get*

Answer 12

\[\frac{a^2-b^2-2abi}{a^2+b^2}=\frac{a^2-b^2}{a^2+b^2}-\frac{2ab}{a^2+b^2}i \]

Answer 13

and for the last one
\[r=\sqrt{(\frac{a^2-b^2}{a^2+b^2})^2+(\frac{-2ab}{a^2+b^2})^2}=1\]

Answer 14

and
\[\theta=\tan ^{-1} (\frac{-2ab}{a^2-b^2})=-\tan ^{-1} (\frac{2ab}{a^2-b^2})\]

Answer 15

I gt this much: -
\[\tan [\log_e{(\frac{re^{ -i \theta}}{re^{i \theta}})^i]}\]

Answer 16

if a+bi = re^i theta

Answer 17

put them in exponential form after you separate imaginary and real part, and it will be easier

Answer 18

\[ \ln(a + ib) = \ln(a^2 + b^2) + i \arctan \left (\frac b a \right ) --(1)\]
\[ \ln(a - ib) = \ln(a^2 + b^2) - i \arctan \left (\frac b a \right ) -- (2)\]
\[ (1) -(2) = 2i\arctan \left (\frac b a \right ) -- (3)\]

Answer 19

So i get
\[\tan [\log_e {\frac{e^\theta}{e^{- \theta}}}]\]

Answer 20

from my way

Answer 21

can plz anyone do from my way?

Answer 22

oh ya i gt this
\[\tan( 2 \theta)\]& \[\theta = \tan ^{-1}(\frac{b}{a}.)\]

Answer 23

now what?

Answer 24

sorry, it was (2) - (1) =
\[ (2) -(1) = -2i\arctan \left (\frac b a \right ) -- (3)\]
\[ (3) \times i = 2\arctan \left (\frac b a \right )\]
\[ \tan \left( 2\arctan \left (\frac b a \right ) \right) = \frac{2ab}{a^2 - b^2}\]

Answer 25

@experimentX Plz plz will u explain my way further:)

Answer 26

first step:
\[ \ln \left ( a - ib \over a + ib \right) = \ln( a - ib) - \ln(a + ib) \]
now follow from what i posted first.

Answer 27

\[\tan(2\theta)=\frac{2tan \theta}{1-tan^2 \theta} \\ \theta=tan^{-1} \frac{b}{a} \\ so \ \ \frac{2tan \theta}{1-tan^2 \theta}=\frac{2b/a}{1-b^2/a^2}=\frac{2ab}{a^-b^2}\]

Answer 28

\[ \tan 2 \theta = \frac{2\tan\theta }{1 - \tan^2\theta }\]

Answer 29

Oh ya i gt it:)
thanx:)

Answer 30

to alll:_)

Answer 31

I mean:)

Answer 32

yw