The least positive integral value of n for which (1+i/1-i) to the power n=1 is (a)2 (b)4 (c)3 (d) none of them
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What is the least positive integral value of for which ?
Let's rationalize the given fraction.
It simplifies to .
Now, the question given to us is, "what is the value of such that ?"
We know that is the definition of the imaginary unit.
If we take to the power of 3
If we take to the power of 4
So, the least positive integral value of is 4. Option (B) is right.
What is the least positive integer that has two equal solutions?
The solution is as follows.
Quadratic discriminant is
To have two equal solutions
This provides that
But the last equation cannot be true.
Hence, the least positive integer is 4.
Answered by
47
Step-by-step explanation:
given :
- least positive integral value of n for which (1+i/1-i) to the power n=1
to find :
- least positive integral value of n = ?
solution :
- please check the attached file
extra information :
- i value = √-1
- i value we called as = iota
Attachments:
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