复数问题的最速方向为什么是共轭梯度

对于复数变量的实值函数 f ( z , z ∗ ) f(z ,z^*) f(z,z∗), 其微分可写为：
d f ( z , z ∗ ) = ∂ f ∂ z d z + ∂ f ∂ z ∗ d z ∗ df(z, z^*) = \frac{\partial f}{\partial z}dz + \frac{\partial f}{\partial z^*}dz^* df(z,z∗)=∂z∂f​dz+∂z∗∂f​dz∗

有：
( ∂ f ∂ z ) ∗ = ∂ f ∗ ∂ z ∗ = ∂ f ∂ z ∗ ( f 为 实 值 ) , d z = ( d z ∗ ) ∗ \left(\frac{\partial f}{\partial z}\right)^*= \frac{\partial f^*}{\partial z^*} = \frac{\partial f}{\partial z^*} (f为实值), dz =(dz^*)^* (∂z∂f​)∗=∂z∗∂f∗​=∂z∗∂f​(f为实值),dz=(dz∗)∗

因此：
∂ f ∂ z d z = ( ∂ f ∂ z ∗ ) ∗ ( d z ∗ ) ∗ = ( ∂ f ∂ z ∗ d z ∗ ) ∗ \frac{\partial f}{\partial z}dz= (\frac{\partial f}{\partial z^*})^*(dz^*)^*=(\frac{\partial f}{\partial z^*}dz^*)^* ∂z∂f​dz=(∂z∗∂f​)∗(dz∗)∗=(∂z∗∂f​dz∗)∗

代入得到：
f = 2 ℜ { ∂ f ∂ z ∗ d z ∗ } = 2 ℜ { ( ∂ f ∂ z ∗ ) ∗ d z } f = 2\Re\{\frac{\partial f}{\partial z^*}dz^*\}=2\Re\{(\frac{\partial f}{\partial z^*})^*dz\} f=2ℜ{∂z∗∂f​dz∗}=2ℜ{(∂z∗∂f​)∗dz}

根据柯西-施瓦茨不等式, d z dz dz应取：

d z = ∂ f ∂ z ∗ dz = \frac{\partial f}{\partial z^*} dz=∂z∗∂f​

也即共轭梯度。