If x=9139199127.....∞,y=4134−194127....∞, and z=∑∞r=1(1+i)−r, then arg(x+yz) is equal to
x=913+132+133...∞
x=9131−13
x=912
x=3
y=413−132+133...∞
y=4131+13
y=414
y=212
y=√2
z=1(1+i)1+1(1+i)2+1(1+i)3...∞
Since
it is a G.P with a common ratio of 1(1+i) we get the
sum
as
=11+i1−11+i
=1i
=−i.
Hence
x+yz=3−√2i
tanθ=−√23
θ=tan−1(−√23)
=−tan−1(√23)