i am getting it 6
3 Answers
spidey timon
·2010-10-31 06:53:57
z1=cis2k1Ï€/21
z2=cis2k2Ï€/28
arg(z1)=2k1Ï€/21
arg(z2)=2k2Ï€/28
arg(z1)=arg(z2)
→k1/k2=3/4
k1 k2 0<k1<20
3 4 0<k2<27
6 8
9 12
12 16
15 20
18 24