How Can I show that the imaginary part of the derivative (and thus the first-order Taylor series) of this function is zero?
${\lvert \boldsymbol{v}^H\boldsymbol{a}+b \rvert}^{2}$
where $\boldsymbol{v}$ and $\boldsymbol{a}$ are two $N*1$ vectors and $b$ is a scalar.also $\boldsymbol{v}$ is the variable.
I know that: ${\lvert \boldsymbol{v}^H\boldsymbol{a}+b \rvert}^{2}=(\boldsymbol{v}^H\boldsymbol{a}+b)*(\boldsymbol{v}^H\boldsymbol{a}+b)^H= \boldsymbol{v}^H\boldsymbol{a}\boldsymbol{a}^H\boldsymbol{v}+\boldsymbol{v}^Hb^H\boldsymbol{a}+b\boldsymbol{a}^H\boldsymbol{v}+\lvert b \rvert^2$
and thus the derivative w.r.t. $\boldsymbol{v}$ is (by supposing that $\boldsymbol{v}$ would real):
$2\boldsymbol{a}\boldsymbol{a}^H\boldsymbol{v}+2*b^H\boldsymbol{a}$
but...