Avoid Doing Complex Arithmetic

This is sort of multiplying things out, but I did not multiply everything out.

$(1+i) (1+2i) = 3i-1$

$(1-i)(1-2i) = -3i - 1 = -(3i + 1)$

For the LHS, we have $(3i-1)(3i+1)$

For the RHS, we have $-(3i+1)(1-3i)$ which is the same as $(3i-1)(3i+1)$

Thus, LHS = RHS and I didn't multiply everything out.

Tan(x+y+z)=(tanx+tany+tanz-tanx.tany.tanz)/(1-tanx.tany-tany.tanz-tanz.tanx)=0

X+y+z=kπ, with x,y,z<π/2 this means k=1

Complex numbers are rotations, All rotations in L are positieve, All rotations in R are negatieve and Both yield to a rotation over angle pi.

Also, if L=R, then LL=RR=100 and thus L=+/-R=+/-10.

With rotation over π, this leaves -10 as a sollution: L=R=-10.

This is a placeholder. If no one else has added a non brute force solution by Sept 1st, I will add mine.

Hint: What is $\tan^{-1} 1 + \tan^{-1} 2 + \tan^{-1} 3$ ?

I'm not quite sure how to zoom it in to make the answer a bit more visible, but I hope you can see the solution. I used trig form of complex numbers to prove the equations true.

Open up interactive Python:

1
2

>>> complex(1,1)*complex(1,2)*complex(1,3) - complex(1,-1)*complex(1,-2)*complex(1,-3)
0j

Proving the statement is equivalent to proving the equivalent form:

$\begin{array} { r l c r} &\dfrac{ ( 1 + i)( 1 + 2i) ( 1+3i) }{ ( 1 - i ) ( 1 - 2i) ( 1-3i) } &= &1 \\ \Leftrightarrow &\left( \dfrac ii \cdot \dfrac{1+i}{1-i} \right) \left( \dfrac ii \cdot \dfrac{1+2i}{1-2i} \right) \left( \dfrac ii \cdot \dfrac{1+3i}{1-3i} \right) &= & 1 \\ \Leftrightarrow & \dfrac{i + i^2}{i - i^2} \cdot \dfrac{i + 2i^2}{i - 2i^2} \cdot \dfrac{i + 3i^2}{i - 3i^2} &= & 1 \\ \Leftrightarrow & \dfrac{(i-1)(i-2)(i-3)}{(i+1)(i+2)(i+3)}&= & 1 \end{array}$

Let $f(x) = (x-1)(x-2)(x-3)$ and $g(x) = (x+1)(x+2)(x+3)$ . By Vieta's formula , we can quickly expand these polynomials to get

$\begin{cases} f(x) = x^3 - 6x^2 + 11x - 6 \\ g(x) = x^3 + 6x^2 + 11x + 6 \end{cases} \; \Rightarrow \; \begin{cases} f(i) = i^3 - 6i^2 + 11i - 6= 10i \\ g(i) = i^3 + 6i^2 + 11i + 6 = 10i \end{cases} \; \Rightarrow \; \dfrac{(i-1)(i-2)(i-3)}{(i+1)(i+2)(i+3)} = \dfrac{f(i)}{g(i)} = 1 \; .$

We're done!