Hey spacemenon.
You are right that 1 + i*tan(theta) can only be in the fourth quadrant however theta can be in any quadrant.
Remember that theta can be in 0 to 2*pi even though tan(theta) only has a principle branch of -pi/2 to pi/2.
find argument and modulus of z=1+itan(theta)
Arg z=theta, and modulus(z)= sec (theta)
case1 0<theta<pi/2
case 2: pi/2<theta< pi
case 3: pi<theta<3pi/2
case 4: 3pi/2<theta<pi
here x=1 and y =tan theta so either the complex no lies in the first quadrant or the fouth quadrant so case 2 and 3 shld be invalid ,but in text there is an anwer for each case hw is that possible ,pls share ur ideas
Hey spacemenon.
You are right that 1 + i*tan(theta) can only be in the fourth quadrant however theta can be in any quadrant.
Remember that theta can be in 0 to 2*pi even though tan(theta) only has a principle branch of -pi/2 to pi/2.