find the constants a b if f(z)=x+2ay+i(3x+by) is analytic
Answers
Answer:
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Answer:
implicit assumption is that z=x+yi and I assume that I stands for i . The function is
f(z)=(1+bi)z+z¯2+(a+ci)z−z¯2i
hence
f(z)=1+bi+a+ci2z+1+bi−ai+c2z¯
The function is analytic if and only if it doesn’t depend on z¯ , so you need 1+c+(b−a)i=0 .
If a,b,c are assumed to be real, the condition is c=−1 and a=b .
You could also use Cauchy-Riemann, with u(x,y)=x+ay and v(x,y)=bx+cy , so
∂u∂x=1,∂u∂y=a,∂v∂x=b,∂v∂y=c
getting the same result.
How do you solve ay''+by'+cy=f(x), where a, b, and c are constants?
If f (x+ay, x-ay) = axy, then what is the value of f(x,y)?
Let f be an analytic function in a neighborhood of 0 having a zero at 0 of order 5 and g has a pole of order 3 at 0 , then f(z) /g(z) has zero of what order? What is the order of its pole?
What is f when u=ay-bx and f(u) = (x+ay) /(1+b)?
What is f(x) if f(x)=x+f(1+1x)?
u=x+ay,v=bx+cy
∂u∂x=∂v∂y so c=1,
∂u∂y=−∂v∂x, so b=−a.
It follows that
f(z)=(x+ay)+i(−ax+y)
=(1−ai)(x+iy)
=(1−ai)z,
which is analytic.
The Cauchy_Rieman equatios
u_x = v_y , u_y = - v_x leads to c= 1 and b = - a .Then follows the following
function f(z) = x + ay + i ( y - ax ) with a arbitrary. If you take
x =( z + zcc)/2 and y = ( z - zcc )/2i where zcc is the complex coniugate of z will
follow that f(z) = ( 1 - ia )z which is analytic being df/dzcc =0-
Step-by-step explanation:
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