Question

Solve: -
\[\huge{\color{purple}{(\frac{x+i}{x-i})^n=1.}}\]

Answer 1

Go by rationalizing the denominator by multiplying and dividing by x + i

Answer 2

ok wait a min.
plz

Answer 3

Take your time..

Answer 4

I gt \[(\frac{x^2-1+2x i}{x^2+1})^n=1\]

Answer 5

You can take both the sides nth root..
can you??

Answer 6

i have just studied that
so i can't:(

Answer 7

\[a^n = 1\]

\[\sqrt[n]{a^n} = \sqrt[n]{1}\]

\[\huge \color{green} {a^{n \times \frac{1}{n}} = (1)^{\frac{1}n}}\]

Till here you got it?

Answer 8

ya

Answer 9

I think I am going wrong wait a minute..

Answer 10

ok

Answer 11

see anything power 0 is 1..
so n should be 0 there..

Answer 12

ok

Answer 13

BUT i m nt getting the answer @waterineyes

Answer 14

Yes but tell me for which variable will we solve is it for n or x??

Answer 15

We have to solve for x & Its answer is \[\cot \frac{k \pi}{n}\]where k= 1,2,3.....n

Answer 16

See till here you got it:

\[(\frac{x^2 - 1 + 2i.x}{x^2 + 1})^n = 1\]

Answer 17

ok then?

Answer 18

\[(\frac{x^2 - 1}{x^2 + 1} + \frac{2x}{x^2 + 1}i)^n = 1\]

Ok??

Answer 19

ok:)

Answer 20

Now we have to make a little substitution:
Put:

\[\frac{x^2 - 1}{x^2 + 1} = sinx\]

now can you find cosx from here??

\[cosx = \sqrt{1 - \sin^2x}\]

Find cosx from here..

Answer 21

ya i can do

Answer 22

Solving or I should solve this??

Answer 23

solving........

Answer 24

Take your time..

Answer 25

I gt
\[(\sin x + i \cos x)^n=1\]\[\implies [ \cos (\frac{\pi}{2}-x)+ i \sin (\frac{\pi}{2}-x)]^n=1.\]now wt should i do?

Answer 26

Do you know about the DE MOIVER'S THEOREM??

Answer 27

of course
so i should apply it here.

Answer 28

Yes you should..

Answer 29

\[[\cos n(\frac{\pi}{2}-x)+i \sin n(\frac{\pi}{2}-x)]=1\]

Answer 30

now?

Answer 31

See, complex number comprises of real and imaginary part..
Since on RHS there is no imaginary part so equate real part of LHS with real part of RHS..

Answer 32

bt then i m getting \[x=\frac{\pi}{2}.\]

Answer 33

No no..
We have to use here general formula:

Do not solve just equate and show me what it has become now.

Answer 34

ok
\[\cos n(\frac{\pi}{2}-x)=1\]\[i \sin n(\frac{\pi}{2}-x)=0i=0.\]

Answer 35

Wait a minute..

Answer 36

ok:)

Answer 37

I am not getting the solution that you want but::

The General solution of:
\[\cos \theta = \cos x\] is:

\[\theta = 2k \pi \pm x\]
where k belongs to set of integer..


So, here

\[cosn(\frac{\pi}{2} - x) = \cos0\]
\[n(\frac{\pi}{2} - x) = 2k \pi \pm 0\]

From here calculate x but the answer is not coming right..

Answer 38

ok np i gt ur solution.
but i think u r right:)

Answer 39

thanx for ur gr8 help:)

Answer 40

Thanks and medal for what??
Answer is not coming right that you want..

Answer 41

medal for gr8 help:)

Answer 42

We are very close to it..

Answer 43

ya:)