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.
This happens specifically with differentiable complex functions. I have a kind of intuitive explanation of this phenomenon.
First, as I hope you know, every complex number can be written in polar form, which looks like this
$$r(\cos(\theta) + i\sin(\theta))$$
for some real $r \ge 0$ and $\theta \in \Bbb{R}$. Multiplying complex numbers in polar form is nice and intuitive. It's easy to verify that:
\begin{align*}
&r_1(\cos(\theta_1) + i\sin(\theta_1)) \times r_2(\cos(\theta_2) + i\sin(\theta_2)) \\
= \, &r_1 r_2(\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)).
\end{align*}
What this means is, if we are multiplying a complex number $z$ in general by a specific complex number $r(\cos(\theta) + i\sin(\theta))$, then all we have to do is:
- Scale $z$ by a factor of $r$, and
- Rotate $z$ (counter-clockwise) by $\theta$ radians.
Note that both of these operations preserve lines and don't change angles. If we start with two lines in the complex plane intersecting at an angle $\alpha$, then multiplying every complex number in these lines by $r(\cos(\theta) + i\sin(\theta))$ will result in two (rotated) lines, still intersecting at an angle of $\alpha$.
Now, let's say we have a complex function $f$ which is differentiable at some complex number $z_0$. Essentially what this means is that the function $f$ behaves like its linearisation:
$$L(z) = f'(z_0)z + f(z_0)$$
for $z$ close to $z_0$. All that $L(z)$ does is multiply $z$ by a fixed complex number $f'(z_0)$ (which just scales and rotates), then adds a fixed complex number $f(z_0)$, which just shifts the picture without changing angles. So, $L$ will also not affect angles!
And since $f(z) \approx L(z)$ for $z$ around $z_0$, this means that, if two smooth curves meet at $z_0$ and their tangent lines meet at an angle $\alpha$, then whatever curves $f$ maps them to will have their tangents meet at angle $\alpha$.
So, what about your particular question? You have a complex function $f(a + ib) = u(a, b) + iv(a, b)$, which we will want to be differentiable. You are looking at the curves $u(a, b) = c_1$ and $v(a, b) = c_2$. The curve $u(a, b) = c_1$ is the set of points which will map to the vertical line $\operatorname{Re} z = c_1$. Similarly, $v(a, b) = c_2$ is the curve that will map to the horizontal line $\operatorname{Im} z = c_2$.
So, if the two curves intersect at $z_0$, and their tangents form an angle $\alpha$, then because $f$ is differentiable at $z_0$ (as it is everywhere else), the image of these curves must also form an angle $\alpha$. But, as we discussed, the image of these curves are horizontal and vertical lines, which are perpendicular! So, we must have $\alpha = \pi/2$. That is, if we look very closely at intersections between the curves, they must occur at right-angles.
Best Answer
We want to express $\tan(a+bi)$ in the form $$\tan(a+bi)=A(a,b)+B(a,b)i,$$ the two functions $A(a,b)$ and $B(a,b)$ are what we are looking for.
We have \begin{align*} \tan(a+bi)&=\frac{\sin(a+bi)}{\cos(a+bi)}\\ &\overset{(1)}=\frac{\sin a\cos(bi)+\cos(a)\sin(bi)}{\cos a\cos(bi)-\sin a\sin(bi)}\\ &\overset{(2)}=\frac{\sin a\cosh b+i\cos a\sinh b}{\cos a\cosh b-i\sin a\sinh b}\\ &=\frac{(\cos a\cosh b+i\sin a\sinh b)(\sin a\cosh b+i\cos a\sinh b)}{\cos^2a\cosh^2b+\sin^2a\sinh^2b}\\ &=\frac{\cos a\sin a(\cosh^2b-\sinh^2b)+i(\cos^2a+\sin^2a)\cosh b\sinh b}{\cos^2a\cosh^2b+\sin^2a\sinh^2b}\\ &=\frac{\cos a\sin a+i\cosh b\sinh b}{\cos^2a\cosh^2b+\sin^2a\sinh^2b} \end{align*}
$(1)$: Using sum angle formulas.
$(2)$: Using $\cos iz=\cosh z$ and $\sinh i = i\sinh z$.
We would like to simplify this further, it seems that double argument formulae might help. In the numerator we can use that $\cos a \sin a = \frac{\sin 2a}2$ and $\cosh b \sinh b = \frac{\sinh2b}2$. For the denominator let us try that from the double angle formulas $\cos2x=\cos^2x-\sin^2x=2\cos^2x-1=1-2\sin^2x$ and $\cosh2x=\cosh^2x+\sinh^2x=2\cosh^2x-1=2\sinh^2x+1$ we can get that \begin{align*} \cos^2x&=\frac{1+\cos2x}2\\ \sin^2x&=\frac{1-\cos2x}2\\ \cosh^2x&=\frac{\cosh2x+1}2\\ \sinh^2x&=\frac{\cosh2x-1}2\\ \end{align*} Now we have for the denominator $D$: \begin{align*} D&=\cos^2a\cosh^2b+\sin^2a\sinh^2b\\ &=\frac{1+\cos2a}2\cdot\frac{\cosh2b+1}2+\frac{1-\cos2a}2\cdot\frac{\cosh2b-1}2\\ &=\frac{1+\cos2a+\cosh2b+\cos2a\cosh2b}4+\frac{1+\cos2a-\cosh2b-\cos2a\cosh2b}4\\ &=\frac{\cos2a+\cosh2b}2 \end{align*} So altogether we get that $$\tan(a+bi)=\frac{\cos a\sin a+i\cosh b\sinh b}{\cos^2a\cosh^2b+\sin^2a\sinh^2b} = \frac{\sin2a+i\sinh2b}{\cos2a+\cosh2b}.$$ So for the real and imaginary part we get \begin{align*} A(a,b)&=\frac{\sin2a}{\cos2a+\cosh2b}\\ B(a,b)&=\frac{\sinh2b}{\cos2a+\cosh2b} \end{align*}