Solving for k1 and k2 in Case Analysis of (|k1|)(|k2|)=1 without Quotation Marks

[fk378]

If (|k1|)(|k2|)=1, show k1, k2= +/- 1

I don't know how to show that abs value of k1=abs value k2=1.

 

[e(ho0n3]

I don't know how to show that abs value of k1=abs value k2=1.

You don't need to do that. You have to consider four cases: k1 = 1, k2 = 1; k1 = -1, k2 = 1; etc. For each case, you have to show that the equality holds.

 

[e(ho0n3]

I'm assuming k1 and k2 are integers by the way. Otherwise, the statement is false.