A friend in need...... (1 Viewer)

[kwabon]

guys...need help
two questions

given that z = x + iy write (2z -1) / (1 + z) in the form A + iB where A and B are real.....(easy one, but still have no clue, stuck at a point)

prove that, if A > 0, squareroot (A + iB), where A and B are real, can always be written in the form x + iy where x and y are real. (totally lost in this question)


your help would be extremely appreciated

 

[Timothy.Siu]

given that z = x + iy write (2z -1) / (1 + z) in the form A + iB where A and B are real.....(easy one, but still have no clue, stuck at a point)

u "real"ise the denominator

(2z -1) / (1 + z)=(2z -1) / (1 + z) x (1+z')/(1+z') where z' is the conjugate,
then u can solve it

prove that, if A > 0, squareroot (A + iB), where A and B are real, can always be written in the form x + iy where x and y are real. (totally lost in this question)

suppose z=root(a+ib)

z^2=a+ib
x²-y²+2ixy=a+ib
then x²-y²=a and 2xy=b x=b/2y

then u like solve it lol...well u dont solve it , u just find out its real

 

[kwabon]

given that z = x + iy write (2z -1) / (1 + z) in the form A + iB where A and B are real.....(easy one, but still have no clue, stuck at a point)

u "real"ise the denominator

(2z -1) / (1 + z)=(2z -1) / (1 + z) x (1+z')/(1+z') where z' is the conjugate,
then u can solve it

prove that, if A > 0, squareroot (A + iB), where A and B are real, can always be written in the form x + iy where x and y are real. (totally lost in this question)

suppose z=root(a+ib)

z^2=a+ib
x²-y²+2ixy=a+ib

then x²-y²=a and 2xy=b x=b/2y

then u like solve it lol...well u dont solve it , u just find out its real

thanks for ur help mate, first one, yep i was on the right path

and the second one, i dont think u really answered the question, u have to prove that squareroot of A + iB can be written in the from of x + iy, sorry i just dont realli get what you are saying.

 

[azureus88]

i'll try another proof for the 2nd question:

let A+iB = rcos@ r>0
sqrt(A+iB)
= sqrt(r)[cos(@/2)+isin(@/2)]
= x+iy where x=sqrt(r)[cos(@/2)] and y=sqrt(r)[sin(@/2)] which are both real as the square root of a positive real number is real and so is the cosine and sine of a real number.

 

[kwabon]

i'll try another proof for the 2nd question:

let A+iB = rcos@ r>0
sqrt(A+iB)
= sqrt(r)[cos(@/2)+isin(@/2)]
= x+iy where x=sqrt(r)[cos(@/2)] and y=sqrt(r)[sin(@/2)] which are both real as the square root of a positive real number is real and so is the cosine and sine of a real number.

YAY!!! legend......that makes a little more sense


thanks for helping u guys, especially u azureus

 

[Timothy.Siu]

lol my bad