# complex numbers

• Jul 10th 2013, 06:28 AM
spacemenon
complex numbers
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(Headbang)(Headbang)(Headbang)
• Jul 10th 2013, 08:37 PM
chiro
Re: complex numbers
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.
• Jul 11th 2013, 04:26 AM
spacemenon
Re: complex numbers
But even if that is the case case 2 and case 3 is invalid
• Jul 11th 2013, 04:32 AM
chiro
Re: complex numbers
Remember that you are looking at theta and not tan(theta).