Question Number 2466 by prakash jain last updated on 20/Nov/15

(√(−(1/5) ))−(1/(√(−5)))=?

Answered by Yozzi last updated on 20/Nov/15

(√(1/(−5)))−(1/(√(−5)))=i(√(1/5))−(1/(i(√5))) =i(1/(√5))+(i/(√5))=2(i/(√5)) (1/(√(−5)))=((√1)/(√(−5)))=(√(1/(−5)))⇒(√(1/(−5)))−(1/(√(−5)))=^? 0

Commented byprakash jain last updated on 20/Nov/15

((2i)/(√5)) is correct. 0 is wrong. since (√(−1))≠(1/(√(−1))) (√(−1))=i, (1/(√(−1)))=(1/i)=−i so (√(−(1/5)))=(i/(√5))

Commented byYozzi last updated on 20/Nov/15

I see.

Answered by Rasheed Soomro last updated on 21/Nov/15

(√(−(1/5) ))−(1/(√(−5)))=? (√(−(1/5)))−((√1)/(√(−5))) (√(−(1/5)))−(√(1/(−5))) ∵ ((√a)/(√b))=(√(a/b)) (√(−(1/5)))−(√(−(1/5))) ∵ ((+ve)/(−ve))=−ve =0 ∵a−a=0 Where is logical mistake?

Commented byFilup last updated on 21/Nov/15

It has to do with the laws of complex radicals. e.g. 1≠(√((−1)(−1)))=(√(−1))(√(−1))=i^2 =−1 Which is incorrect (√a)×(√b)≠(√(−a))×(√(−b)) (√a)×(√(−b))≠(√(−a))×(√b) (in some situations) such as: for: (√(−(1/5)))−(1/(√(−5)))⇒(√(−(1/5)))≠(1/(√(−5)))

Commented byRasheed Soomro last updated on 21/Nov/15

THanKs. That means (√(ab))=(√a)(√b) and (√(a/b))=((√a)/(√b)) are restricted. They are aplicable only when a>0 and b>0.

Commented byFilup last updated on 21/Nov/15

Exactly. Here is another example: −(1/(√(−2)))=−(1/(i(√2)))=(i/(√2)) (i^(−1) =−i) Alternitively: −(1/(√(−2)))=(√(((−1)^2 )/(−2)))=(√(1/(−2)))=(√((−1)/2))=(i/(√2)) Here we are allowed to transfer the signs such that (√a)(√(−b))=(√(−a))(√b)

Commented byFilup last updated on 21/Nov/15

Also note this: (√(−a))×(√(−b))≠(√(ab)) Proof (√(−a))=i(√a) (√(−b))=i(√b) ∴(√(−a))(√(−b))=i^2 (√(ab))=−(√(ab)) (√(ab))≠(√(−a))(√(−b)) example 1=(√1)=(√((−1)(−1)))≠(√(−1))(√(−1))=i^2 =−1

Commented byRasheed Soomro last updated on 22/Nov/15

Is (√(−a))(√b) =(√(−ab ))? [ a,b>0]

Answered by Filup last updated on 21/Nov/15

Follwing the laws of complex radicles, both answers of ((2i)/(√5)) and 0 are correct. I belive that we have two solutions. ((2i)/(√5)) is our complex solution (ℑ(z)=(2/(√5))) 0 is our real solution If z=ℜ(z)+ℑ(z) we have: z=0+((2i)/(√5))

Commented byRasheed Soomro last updated on 22/Nov/15

The two ways for achieving ((2i)/(√5)) and 0 are not simultaneous. So these answers can′t be correct at same time.You can′t add them to achieve z