if Im(f ′(z)) =6x(2y−1) and f(0)=3−2i , f(1)=6−5i find f(1+i)?


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Question Number 73964 by arkanmath7@gmail.com last updated on 17/Nov/19

$i f I m (f^{'} (z)) = 6 x (2 y - 1) a n d$ $f (0) = 3 - 2 i, f (1) = 6 - 5 i$ $f i n d f (1 + i) ?$
	Answered by mind is power last updated on 17/Nov/19

	$f^{'} (z) is \frac{df}{dz} . . ?$
	Commented by arkanmath7@gmail.com last updated on 17/Nov/19

	$y e s$
	Answered by mind is power last updated on 17/Nov/19

	$f (x + iy) = p (x, y) + iQ (x, y)$ $f^{'} (z) = \frac{\partial}{\partial z} (f (z))$ $\frac{\partial}{\partial z} = \frac{1}{2} (\frac{\partial}{\partial x} - i \frac{\partial}{\partial y}) = \frac{1}{2} (\frac{\partial p}{\partial x} + \frac{\partial Q}{\partial p}) - \frac{i}{2} (\frac{\partial p}{\partial y} - \frac{\partial Q}{\partial x})$ $Im (f^{'} (z)) = {imf}^{'} (x + iy) = \frac{1}{2} (\frac{\partial Q}{\partial x} - \frac{\partial p}{\partial y}) = 6 x (2 y - 1) . . . . . E$ $f holomorphic \Rightarrow \frac{\partial p}{\partial x} + i \frac{\partial Q}{\partial x} = - i \frac{\partial p}{\partial y} + \frac{\partial Q}{\partial y}$ $\Rightarrow \frac{\partial Q}{\partial x} = - \frac{\partial P}{\partial y} \Rightarrow E \frac{\partial Q}{\partial x} = 12 xy - 6 x \Rightarrow Q (x, y) = {6 x}^{2} y - {3 x}^{2} + H (y)$ $\Rightarrow P (x, y) = - {6 xy}^{2} + 6 xy + f (x)$ $\frac{\partial p}{\partial x} = \frac{\partial Q}{\partial y} \Leftrightarrow - {6 y}^{2} + 6 y + f^{'} (x) = {6 x}^{2} + H^{'} (y)$ $\Rightarrow f^{'} (x) = {6 x}^{2} \Rightarrow f (x) = {2 x}^{3} + c$ $H^{'} (y) = - {6 y}^{2} + 6 y \Rightarrow H (y) = - {2 y}^{3} + {3 y}^{2} + d$ $\Rightarrow f (x + iy) = (- {6 xy}^{2} + 6 xy + {2 x}^{3} + c) + i ({6 x}^{2} y - {3 x}^{2} - {2 y}^{3} + {3 y}^{2} + d)$ $f (1) = 6 - 5 i,$ $f (0) = 3 - 2 i \Rightarrow c = 3, d = - 2$ $f (1) = (2 + 3) + i (- 3 - 2) = 5 - 5 i$ $f {did}^{'} nt exist withe this conditions$
	Commented by arkanmath7@gmail.com last updated on 18/Nov/19

	$t h m x s i r$
	Commented by mind is power last updated on 18/Nov/19

	$y^{'} r e w e l c o m$
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