Question

If a+ib = c+i / c-i, where c is real, prove that a 2 + b 2 =1 and b/a = 2c / c 2 -1.

Answer 1

Answer:

A+ib=c+i/c-i -------- 1

On taking conjugate of eq. 1

A-ib =c-i/c+i. -------- 2

Now,multiply eq 1 and eq. 2

Then we get,

(A+ib)×(A-ib) = (c+i)×(c-i)/(c-i)×(c+i)

A^2+b^2 =c^2+1/c^2+1

Since,(i^2=-1) and

(a+b)(a-b)=a^2-b^2. (^= power)

So, c+1/c+1 will cancel out

And answer is,

A^2+b^2 =1

Hence proved.

Now ,

A + iB = C + i / C- i

Converting it into standard form

(C+i) (c+i)/(c-i)(c+i)

C^2+i^2+2ci/c^2+1

(c^2-1/c^2+1) + i (2c/c^2+1)

Now comparing it with a+ib

We get a= c^2-1/c^2+1 ,

b= 2c/c^2+1

now by dividing b by a we will get,

2c/c^2-1 (hence proved )

Step-by-step explanation: