CSIR NET MATHEMATICS| How to solve the problem of Complex Analysis by Counter Examples?(June2018)

In this video, we are going to discuss the problem of complex analysis from the CSIR NET June 2018. The problem is, consider the map, pi, defined from complex plane minus 1 to complex plane given by, pi of z is equal to 1 plus z divided by 1 minus z. Then which of the following is true? Option 1, pi of z belongs to T, such that, modulus z less than 1, subset of z belongs C belongs to C. Modulus Z less than 1. So, unit open disk, its image is also into the unit open disk. Second option, pi of z belongs to c such that real part of z less than 0, subset of z belongs to c such that real part of z less than 0. This real part of z less than 0 represent the left half plane. So, second option says that image of left half plane is contained in the left half plane. Third option, pi on to. And the fourth option, pi of c minus 1 is equal to c minus minus 1. Now, we are using the counter example. So, first option, we assume z is equal to 0.5. This is in the unit open disk. Now, we are going to find the image of z is equal to 0.5. So, pi of 0.5 is equal to 1 plus z. So, z is equal to 0.5. 1 plus 0.5 divided by 1 minus 0.5. is equal to 1.5 divided by 0.5 and has value 3, which is not less than 1. Therefore, modulus 0.5 is equal to 3. So, this image does not belong to the unit open disk. Therefore, the first option is wrong. Now, we are going to check the second option. Now, with the counter example, that is equal to minus 0.5. So, this minus 0.5 is less. This real part is minus 0.5 which is less than 0. So, it is lie in the left half plane and its image 5 minus 0.5 is equal to 1 plus minus 0.5 divided by 1 minus minus 0.5. So, 0.5 divided by 1.5, 1 by 3 which is positive. Therefore, its image does not belongs to the left half plane. So, the second option that the image of left half plane containing Left half plane is wrong. Now, the third option to check onto. So, we assume that f of z is equal to value some w. We have function 1 plus z divided by 1 minus z is equal to w divided by 1. Now, using dividend component method, 1 plus z plus denominator 1 minus z divided by 1 plus z minus denominator 1 minus z is equal to w plus 1. So, we have numerator 2 divided by 2z is equal to w plus 1 divided by w minus 1. Therefore, z is equal to w minus 1 divided by w plus 1. Now, this is valid if w not equal to minus 1. So, we cannot get any value corresponding to w minus 1 in the domain space c minus 1. The image is pi of c minus 1 is equal to c minus minus 1. So which is not onto first of all. So third option is cancel and the fourth option is correct. Pi of c minus 1 is equal to c minus minus 1. Thanks for watching. For more videos like subscribe. Press the bell icon of Next Gen Maths