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Complex Function ( Limit, Analytic Function)

Fz is a funct...

Question

$F (z)$ is a function of the complex variable $z = x + i y$ given by
$F (z) = i z + k R e (z) + i I m (z)$
For what value of $k$ will $F (z)$ satisfy the Cauchy-Riemann equations?

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Solution

The correct option is B $1$
$F (z) = i z + k R e (z) + i I_{m} (z)$
$= i (x + i y) + k (x) + i (y)$
$= (k x - y) + i (x + y)$
$= u + i v$
So $u = k x - y$ and $v = x + y$
By C-R equation:
$u_{x} = v_{y}$ and $u_{y} = - v_{x}$
$(R = 1) = R (1 - = - 1)$
Hence $k = 1$

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F(z) is a function of the complex variable z=x+iy given by F(z)=iz+kRe(z)+i Im (z) For what value of k will F(z) satisfy the Cauchy-Riemann equations?

The correct option is B 1F(z)=iz+kRe(z)+iIm(z) =i(x+iy)+k(x)+i(y) =(kx−y)+i(x+y) =u+iv So u=kx−y and v=x+y By C-R equation: ux=vy and uy=−vx (R=1)=R(1−=−1) Hence k=1

$F (z)$ is a function of the complex variable $z = x + i y$ given by
$F (z) = i z + k R e (z) + i I m (z)$
For what value of $k$ will $F (z)$ satisfy the Cauchy-Riemann equations?

The correct option is B $1$
$F (z) = i z + k R e (z) + i I_{m} (z)$
$= i (x + i y) + k (x) + i (y)$
$= (k x - y) + i (x + y)$
$= u + i v$
So $u = k x - y$ and $v = x + y$
By C-R equation:
$u_{x} = v_{y}$ and $u_{y} = - v_{x}$
$(R = 1) = R (1 - = - 1)$
Hence $k = 1$