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Let $z=x+iy$, $x$ and $y$ are real numbers. $f(z)=6x-4y+i(ax+by)$, $a$ and $b$ are real numbers. If $f(z)$ is differentiable, find $a$ and $b$.

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Are you familiar with the Cauchy Riemann Equations? –  JSchlather Oct 21 '12 at 8:29
    
You need to check if $f(z+h)-f(z)=w\cdot h+o(h)$. Calculate and simplify the left hand side with $z=x+iy$ and $h=u+iv$. –  Hagen von Eitzen Oct 21 '12 at 8:31
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If a complex fuction $f(z)=u(z)+iv(z)$ is differentiable, it's real and imaginary parts should satisfy Cauchy-Riemann condition: $$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$$ and $$\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$$ so in your case $\frac{\partial u}{\partial x} = 6$,$\frac{\partial v}{\partial y} = b$, $\frac{\partial u}{\partial y} = -4$ ,$-\frac{\partial v}{\partial x} = a$. It gives: $$a =4,b=6$$

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If $f$ is differentiable as a complex function, it must fulfill $$ \begin{eqnarray*} \frac{\partial \Re(f(x+iy))}{\partial x} &=& \frac{\partial \Im(f(x+iy))}{\partial y} \\ \frac{\partial \Re(f(x+iy))}{\partial y} &=& -\frac{\partial \Im(f(x+iy))}{\partial x} \end{eqnarray*} $$ where $\Re(z)$ denotes the real part and $\Im(z)$ the imaginary part of a $z \in \mathbb{C}$. Those equations are called the Cauchy-Riemann differential equations.

If you compute these derivatives, you should be able to determine which values $a$ and $b$ must take to have your $f$ fulfill these equations.

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