CSIR NET MATHEMATICS| Complex Analysis | Easiest Problem of DEC 2018-Anyone can solve (Marks 4.75)

In this video we are going to discuss the example of complex analysis from CSIRnet December 2018. Example 86 from the booklet C which is of Marks 4.75 and this is one of the easiest problem of CSIRnet December 2018. Let h denote the upper half plane that is h is equal to x plus i y, y greater than zero for z belongs to H. Which of the following are true? First option 1 by z belongs to H. Second option 1 upon z square belongs to H. Third option minus z upon z plus 1 belongs to H. Fourth option z divided by 2z plus 1 belongs to H. Now we are given h is a upper half plane. That means y is greater than 0. So, geometrically it contains this upper part of the complex plane where y is greater than 0, x may be negative or may be positive. Now, first option, so h is equal to x plus iy with the condition y greater than 0, that means any element of h, the imaginary part y should be positive. If it is not positive, then we can say that that element does not belong to h. So first option 1 by Z. So 1 by Z is equal to 1 upon X plus IY multiply by and divide by its conjugate X minus IY. So its conjugate is X minus IY divided by X minus IY. Now in the numerator 1 into X minus IY is X minus IY and the denominator is X square minus I square Y square minus I square is equal to 1. So it is X square plus Y square. Now separating the real and imaginary parts so x upon x square plus y square is real part and the imaginary part is minus y upon x square plus y square. As z belongs to H therefore y is always greater than 0 and because of that minus y upon x square plus y square is less than 0 as the denominator is a sum of square of two numbers and hence The imaginary part is negative therefore 1 by z does not belong to H and hence the option 1 is wrong. Now similarly option 2 1 upon z square. z is equal to x plus i y so denominator is x plus i y whole square. Now expand this bracket then we have x square minus y square plus 2 i x y multiply and divide by the conjugate of denominator So, x square minus y square minus i 2xy divided by x square minus y square minus i 2xy. In the numerator, 1 multiplied x square minus y square minus i 2xy and the denominator we have x minus y whole square plus 4 times x square y square. Here, we are using the property of i, i square equal to minus 1. Now, separating the real and imaginary part. So, imaginary part is It is minus 2xy. Divide by x minus y square whole square plus 4x square y square. z belongs to H, y is always positive. Suppose x is also positive then 2xy is positive but minus 2xy is negative. So in that case this imaginary part will be negative and therefore 1 upon z square does not belong to H if x and y both are positive numbers. So, option 2 is also wrong. Now, similarly we are going to check the option 3 where we have minus z upon z plus 1. So, z replaced by x plus iy. So, x plus iy plus 1. In the denominator, we are gathering the real part. So, x plus 1 plus iy. Then, multiply its conjugate which is x plus 1 minus iy. Then, real part is minus x multiply with the x plus 1. And for second term of the real part, Iy multiply with Iy, so minus iy multiply to minus iy, so plus i square y square, and plus i square is equal to minus 1, so second term of the real part is minus y square. Plus, for the imaginary part, minus x multiplied by minus iy, so we have plus ixy. For the second term of imaginary part, minus iy multiply with x plus 1 so we have minus y times x plus 1 in the imaginary part if we open the second bracket so plus xy cancel with the minus xy and we have minus 1. So second part will give the minus 1 and therefore the imaginary part is minus 1 upon x plus 1 whole square plus y square. As the denominator is sum of the square of real numbers so it is always positive So, minus 1 upon x plus 1 whole square plus y square is negative. So, imaginary part is negative and because of that minus z upon 1 plus z does not belong to h as the condition is imaginary part should be positive and hence the option 3 is wrong. Now, for the fourth option, we have z divided by 2z plus 1 where z is equal to x plus iy. Now, in the denominator gathering the real parts, we have 2x plus 1 plus i times 2y. Now, multiply and divide by the conjugate of denominator, which is 2x plus 1 minus i times 2i. So, in the numerator, the real part is x into 2x plus 1 and the second term of real part is i y multiply the minus. i 2y so which is minus i square 2y square is equal to minus i square is plus 1 therefore the second term of real part is 2y square plus for the imaginary part x multiply the minus i 2y which is i times minus 2xy and for the second term of imaginary part iy multiply the 2x plus 1 It is y times 2x plus 1. Now separating the real and imaginary parts. So, in the imaginary part, minus 2xy cancel with the plus 2xy and we have only 1 into y which is y divided by 2x plus 1 whole square plus 4y square. As the condition for h, z belongs to h, the y is always positive, the imaginary part. So, y upon 2x plus 1 whole square 4y square is also positive and therefore, the imaginary part is positive. Hence, by the condition of H, Z upon 2Z plus 1 belongs to H and hence, the fourth option is correct. So, this example, only fourth option is correct. So, this is one of the easiest problem of December 2018. Thanks for watching. For more videos, like, subscribe, press the bell icon of NextGenMath.