# Thread: Complex arctan

1. ## Complex arctan

I found these two questions in a textbook and they are driving me nuts.

How can you derive from $\displaystyle (1+i)(2+i)(3+i)=10i$ that $\displaystyle \arctan1+\arctan\frac{1}{2}+\arctan\frac{1}{3}=\fr ac{\pi}{2}$?

Similarly, how can you with the help of $\displaystyle (1-2i)^4 = -7+24i$ show that $\displaystyle 4\arctan2=\pi+\arctan\frac{24}{7}$

2. Originally Posted by Mondreus
I found these two questions in a textbook and they are driving me nuts.

How can you derive from $\displaystyle (1+i)(2+i)(3+i)=10i$ that $\displaystyle \arctan1+\arctan\frac{1}{2}+\arctan\frac{1}{3}=\fr ac{\pi}{2}$?
1+ i can be written, in polar form, as $\displaystyle \sqrt{2}(cos(\theta_1)+ i sin(\theta_1)$ where $\displaystyle tan(\theta_1)= 1/1= 1$ (and, of course, $\displaystyle \theta_1= \pi/4$). 2+ i can be written as $\displaystyle \sqrt{5}(cos(\theta_2)+ i sin(\theta_2))$ where $\displaystyle tan(\theta_2)= 1/2$. 3+ i can be written as $\displaystyle \sqrt{10}(cos(\theta_3)+ i sin(\theta_3)$ where $\displaystyle tan(\theta_3)= 1/3$.

By DeMoivre's formula, their product is $\displaystyle (\sqrt{2})(\sqrt{5})\sqrt{10}(cos(\theta_1+ \theta_2+ \theta_3)+ i sin(\theta_ 3+ \theta_2+ \theta_3))$.

The fact that the product is 10 i tells you that $\displaystyle cos(\theta_1+ \theta_2+ \theta_3)= 0$ and $\displaystyle cos(\theta_1+ \theta_2+ \theta_3)= 1$ or that $\displaystyle \theta_1+ \theta_2+ \theta_3= \pi/2$.

Similarly, how can you with the help of $\displaystyle (1-2i)^4 = -7+24i$ show that $\displaystyle 4\arctan2=\pi+\arctan\frac{24}{7}$
$\displaystyle 1- 2i= \sqrt{5}(cos(\theta+ i sin(\theta)$ where $\displaystyle tan(\theta_1)= 2$ and, again by DeMoivre's formula, that is equal to $\displaystyle 25(cos(4\theta)+ i sin(4\theta))$. -7+ 24i can be written as $\displaystyle 25 (cos(\phi)+ i sin(\phi))$ where $\displaystyle tan(\phi)= 24/7$.

3. What about the $\displaystyle \pi$ part in the last problem?

4. Consider which quadrant you are in.

5. Originally Posted by HallsofIvy
1+ i can be written, in polar form, as $\displaystyle \sqrt{2}(cos(\theta_1)+ i sin(\theta_1)$ where $\displaystyle tan(\theta_1)= 1/1= 1$ (and, of course, $\displaystyle \theta_1= \pi/4$). 2+ i can be written as $\displaystyle \sqrt{5}(cos(\theta_2)+ i sin(\theta_2))$ where $\displaystyle tan(\theta_2)= 1/2$. 3+ i can be written as $\displaystyle \sqrt{10}(cos(\theta_3)+ i sin(\theta_3)$ where $\displaystyle tan(\theta_3)= 1/3$.

By DeMoivre's formula, their product is $\displaystyle (\sqrt{2})(\sqrt{5})\sqrt{10}(cos(\theta_1+ \theta_2+ \theta_3)+ i sin(\theta_ 3+ \theta_2+ \theta_3))$.

The fact that the product is 10 i tells you that $\displaystyle cos(\theta_1+ \theta_2+ \theta_3)= 0$ and $\displaystyle cos(\theta_1+ \theta_2+ \theta_3)= 1$ or that $\displaystyle \theta_1+ \theta_2+ \theta_3= \pi/2$.

$\displaystyle 1- 2i= \sqrt{5}(cos(\theta+ i sin(\theta)$ where $\displaystyle tan(\theta_1)= 2$ and, again by DeMoivre's formula, that is equal to $\displaystyle 25(cos(4\theta)+ i sin(4\theta))$. -7+ 24i can be written as $\displaystyle 25 (cos(\phi)+ i sin(\phi))$ where $\displaystyle tan(\phi)= 24/7$.

Wow this problem is confusing. It took me a little bit to figure it out.

HallsofIvy you have a typo when you said

$\displaystyle cos(\theta_1+ \theta_2+ \theta_3)= 1$

it should be

$\displaystyle sin(\theta_1+ \theta_2+ \theta_3)= 1$

Thanks for working it out though. It really did help me figure out my mistake.

I kept working out

$\displaystyle 4\arctan(-2)=\pi+\arctan\bigg(-\frac{24}{7}\bigg)$

Since the original problem was in the second quad. Didn't notice the tan had been changed to positive value changing the problem.

6. Originally Posted by 11rdc11
Wow this problem is confusing. It took me a little bit to figure it out.

HallsofIvy you have a typo when you said

$\displaystyle cos(\theta_1+ \theta_2+ \theta_3)= 1$

it should be

$\displaystyle sin(\theta_1+ \theta_2+ \theta_3)= 1$
Yes, thanks for the correction.

Thanks for working it out though. It really did help me figure out my mistake.

I kept working out

$\displaystyle 4\arctan(-2)=\pi+\arctan\bigg(-\frac{24}{7}\bigg)$

Since the original problem was in the second quad. Didn't notice the tan had been changed to positive value changing the problem.