interesting and challenging ds

IMO A

We know that z^n = 1. We are asked for the value of z.

(I) n is a non-zero integer.

Here n could take positive or negative integers. In any case, z will always be 1 to satisfy z^n = 1. z cannot take 0 or -1 or any number other than 1. Hence, only 1 value. Sufficient.

(II) z>0

Here z can take any positive integer. n could only be 0. But we are not bothered about n.
1^0 = 1
2^0 = 1
.
.
You get multiple z values. Insufficient.

stmnt (1) is not suffcient .. when n is a even integer u can have z = -1 or 1 ,as Z^even =1
z = 1 only when n is odd integer

Oh yes! Forgot to consider that case. Thanks for the catch

IMO C

Clearly Statements 1 and 2 are not enough on their own.

But when when you combine both, you can conclude that the only way z^n = 1 when z>0 and n is not 0, is when z=1.

Hence we can find the value of z by combining both statements and therefore IMO C

shouldn't it be (B), we're told z^n = 1 from stmt II z>0, z^n = 1 only if z = 1 coz 1^anything is 1

shouldn't it be (B), we're told z^n = 1 from stmt II z>0, z^n = 1 only if z = 1 coz 1^anything is 1

Nopes. Even if z>o, z can take multiple values if you take n as 0 (which you can because Statement 2 alone does not put any conditions on the value of n). So if n=0, even 1^0 = 1, 2^0 = 1, 3^0 = 1.....and so on. Hence z can take on multiple values if you take only Statement 2 into account.

It's only when you combine Statement 2 with the first one, that you can conclude that the only possible value that z can take is 1.

Hence C