Question

hello is there anyone to help me solve qustions related to complex number class 11 reply

Answer 1

What is the question Kapuria?

Answer 2

the qustion is simlify i35+1/i35

Answer 3

is it : \(\large\cfrac{i^{35} + 1}{i^{35}} \) ?

Answer 4

heh noo itss written seprately i35 + 1/i35

Answer 5

Ok. \(\large i^{35} + \cfrac{1}{i^{35}} \) .

Can you find the value for \(\large i^{35} \) ?

Answer 6

Divide 35 by 4. You get remainder as 3 . So \(\large i^{35} = i^3 \) . As \(\large i^3 = i * i * i = -1 *i = -i \)

So : \(\large i^{35} = -i\)

Answer 7

Plug-in the value of \(\large i^{35} \) and solve it further.

Answer 8

well here is a simple thing... use a common denominator

\[\frac{i^{35} \times i^{35}}{i^{35} }+ \frac{1}{i^{35}} \]

so you have

\[\frac{i^{70} + 1}{i^{35}}\]

just need to evaluate

\[i^{70}\]

its not to difficult

Answer 9

so when you simplify it you get

\[\frac{(i^2)^{35} + 1}{i^{35}}\]

substitute

\[i^2 = -1\]

and work from there

Answer 10

Will it be better to let it be in the form of \(\large i^{35} + \cfrac{1}{i^{35}} \)

\(i^{35} = -i \)

It will be easy to move on and simplify.

Answer 11

tell me the final answer i hve then i34 i + 1/i34 i then wht is the nxt step after thiss

Answer 12

@kapuria Please look the above posts and try again.

Answer 13

kkkkkkk

Answer 14

ans is coming -i is this corrct

Answer 15

Hmm, no! See, \(\large i^{35} = -i \) , right?

Answer 16

yess then

Answer 17

I have : \(\large i^{35} + \cfrac{1}{i^{35}} = -i + \cfrac{1}{-i} = -i + (-i) = - i - i = -2i\)

Answer 18

kkkkkkk got it thnksss

Answer 19

You're welcome. :)

Answer 20

if u donot mind may i ask u 4 more qustions also

Answer 21

Oh k wait @kapuria

Answer 22

First of all , I apologize for my mistake. The answer given by me is wrong.

Answer 23

See, I had : \(-i + \cfrac{1}{-i} \) ,

\(-i + \cfrac{1}{-i} \times \cfrac{-i}{-i} \)

[What I did was just, I multiplied \(\cfrac{1}{-i} \) by \(\cfrac{-i}{-i}\) ]

Answer 24

Now, \(-i + \cfrac{-i}{-i * -i } \) = \(-i + \cfrac{ -i}{(i^2)} \)

\(-i + \cfrac{-i}{-1} \) = \(-i + i = 0\)

Answer 25

So the answer is 0. [Sorry for my mistake again]

Answer 26

kkkkkkk r u sure now

Answer 27

Yep . Very sure. :)

Answer 28

the qustion is find value of x and y

(1+i) y ka squre + (6+i) = (2+i)x

solve plzz

Answer 29

Please post this as a new question. Some one will surely help you.