Ideally what you want to have at the end is a nice complex number of the form :
$$z = x+iy$$
Unfortunately as you have shown, complex numbers aren't always written this way... Being able to write them under an exponential form is a great way to evaluate them for a few reasons.
The first reason I see is that it is very easy to take their conjugate form, simply change the sign of the argument.
Another good reason is that $re^{i\theta} = r\cos(\theta) + ir\sin(\theta)$ making it easy to transform into a nice form.
They are very easy to multiply and divide with each another.
One problem though : you can't add them simply.
So when evaluating horrific complex numbers here are my advice:
- If you need to add or subtract complex numbers of the form $x + iy$, then just use the normal formula.
- If you have expression you want to add but aren't of the form $x+iy$, you should normally be able to write them as a polar form and then rewrite them as $x+iy$ using Euler's formula.
- If you want to divide, multiply or take a power of two complex numbers, always use exponential forms, it is the easiest way.
With all of this you should (slowly) be able to evaluate complex numbers written using elementary operations. As you can see, the exponential form is used very often except when adding or subtracting. In the end you should always be able to write it under the form $x+iy$ where the imaginary and real parts are explicit.
It's really just a question of the definition of the derivative. If $z=x+yi,$ $f(z)=u(x,y)+iv(x,y)$ can be any pair of functions $u,v.$
But if $f$ is differentiable, then:
$$f'(z)=\lim_{h\to 0}\frac{f(z+h)-f(z)}{h}\tag{1}$$
then $h$ can approach $0$ in many different ways, since $h$ is complex.
For example, you can have $h\to 0$ on the real line. Then: $$f'(z)=\frac{\partial u}{\partial x} +i\frac{\partial v}{\partial x}$$
But if $h\to 0$ along the imaginary part, then:
$$\begin{align}f'(z)&=\frac{1}{i}\left(\frac{\partial u}{\partial y}+i\frac{\partial v}{\partial y}\right)\\
&=\frac{\partial v}{\partial y}-i\frac{\partial u}{\partial y}
\end{align}$$
So for the limit to be independent of any path you take $h\to 0$ you must have at minimum that $$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y},\\\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}\tag{2}$$
So for (1) to be true, we need $u,v$ to satisfy the differential equations in (2).
It turns out that $(2)$ is enough to ensure that $(1)$ converges to a single value, but that is not 100% obvious.
The equations in (2) are called the Cauchy-Riemann equations.
Another way of looking at it is, given a function $f:\mathbb R^2\to\mathbb R^2$ mapping $\begin{pmatrix}x\\y\end{pmatrix}\mapsto \begin{pmatrix}u(x,y)\\v(x,y)\end{pmatrix}$ there is a matrix derivative standard from multi-variable calculus:
$$Df\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}\frac{\partial u}{\partial x}&\frac{\partial u}{\partial y}\\\frac{\partial v}{\partial x}&\frac{\partial v}{\partial y}\end{pmatrix}\tag{3}$$
For small vectors $$\mathbf h=\begin{pmatrix}h_1\\h_2\end{pmatrix}$$ you get $f\left(\begin{pmatrix}x\\y\end{pmatrix}+\mathbf h\right)\approx f\begin{pmatrix}x\\y\end{pmatrix}+Df\begin{pmatrix}x\\y\end{pmatrix}\mathbf h.$
In particular, $Df$ is in some sense the "best" matrix, $\mathbf A,$ for estimating $f(\mathbf v+\mathbf h)\approx f(\mathbf v)+\mathbf A\mathbf h.$
Now, these matrices are not complex numbers. But an interesting fact is that the set of matrices of the form:
$$\begin{pmatrix}a&-b\\b&a\end{pmatrix}\tag{4}$$
are a ring isomorphic to the ring of complex numbers. Specifically, the above matrix corresponds to $a+bi.$
We also have that:
$$\begin{pmatrix}a&-b\\b&a\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=
\begin{pmatrix}ax-by\\bx+ay\end{pmatrix}$$
compare that with:
$$(a+bi)(x+yi)=(ax-by)+(ay+bx)i.$$
So these matrices (4) act on $(x,y)^T$ the same way that $a+bi$ acts on $x+yi$ by multiplication.
The Cauchy-Riemann equations (2) just mean that $Df\begin{pmatrix}x\\y\end{pmatrix}$ is an example of (4) - that is, when the Cauchy-Riemann equations are true for $u,v$ then the multi-variate derivative (3) can be thought of as a complex number.
So we see that when we satisfy Cauchy-Riemann, $Df\begin{pmatrix}x\\y\end{pmatrix}\cdot\mathbf h$ can be seen as multiplication of complex numbers, $f'(z)$ and $h=h_1+h_2i.$ Then you have:
$$f(z+h)\approx f(z)+f'(z)h.$$
where $f'(z)$ is not just the best estimating complex number for this approximation, but also $f'(z)$ is the best linear operation on $h$ for this estimation.
So complex analysis is taking the vector function and asking, $f$ "when does it make sense to think of the derivative of the $\mathbb R^2\to\mathbb R^2$ as a complex number?" That is exactly when Cauchy-Riemann is true.
In the general case $f:\mathbb R^2\to\mathbb R^2,$ we can't really take the second derivative and get an estimate $f(z+h)\approx f(z)+Df(z)\cdot h +\frac{1}{2}D^2f(z)\cdot h^2+\cdots.$ We can't get easy equivalents to power series approximations of $f.$
But when $Df$ satisfies Cauchy-Riemann, we can think if $Df$ as a complex-valued function.
So complex analysis is a subset of the real analysis of functions $\mathbb R^2\to\mathbb R^2$ such that the derivative matrix $Df$ can be thought of as a complex number. This set of functions turns out to have a lot of seemingly magical properties.
This complex differentiability turns out to be fairly strong property on the functions we study. The niceness of the Cauchy-Riemann equations gives up some truly lovely results.
Best Answer
For the modulus to be $1,$ you only need the product $Bf$ to be $\pm1,$ for since $$|1+iBf|=1,$$ it follows that $$(1+iBf)(1-iBf)=1+(Bf)^2=1,$$ and the result I claimed follows.
Of course I assumed $B$ and $f$ are real, since you do not say anything about them. Otherwise what I say above needs to be modified to be true.
OK, you have specified that $f(\lambda)$ is complex valued. Thus, if you write $f=a+ib,$ where $a=a(\lambda),\,\, b=b(\lambda).$ Then we have that $$1+iBf=1+iB(a+ib)=1-Bb+iBa.$$
For its modulus to be $1,$ we must have $$(1-Bb+iBa)(1-Bb-iBa)=1,$$ or $$(1-Bb)^2+(Ba)^2=1.$$ This gives $$B=\frac{2\Im f(\lambda)}{|f(\lambda)|^2}.$$