1. ## Complex arctan

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

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

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

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

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

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

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

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

3. What about the $\pi$ 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 $\sqrt{2}(cos(\theta_1)+ i sin(\theta_1)$ where $tan(\theta_1)= 1/1= 1$ (and, of course, $\theta_1= \pi/4$). 2+ i can be written as $\sqrt{5}(cos(\theta_2)+ i sin(\theta_2))$ where $tan(\theta_2)= 1/2$. 3+ i can be written as $\sqrt{10}(cos(\theta_3)+ i sin(\theta_3)$ where $tan(\theta_3)= 1/3$.

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

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

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

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

HallsofIvy you have a typo when you said

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

it should be

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

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

I kept working out

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

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

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

it should be

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

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

I kept working out

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

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