# Complex arctan

• December 26th 2009, 03:00 AM
Mondreus
Complex arctan
I found these two questions in a textbook and they are driving me nuts. (Thinking)

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 $4\arctan2=\pi+\arctan\frac{24}{7}$
• December 26th 2009, 03:47 AM
HallsofIvy
Quote:

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

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 $\theta_1+ \theta_2+ \theta_3= \pi/2$.

Quote:

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 $tan(\phi)= 24/7$.
• December 26th 2009, 04:21 AM
Mondreus
What about the $\pi$ part in the last problem?
• December 26th 2009, 06:08 AM
HallsofIvy
Consider which quadrant you are in.
• December 26th 2009, 03:25 PM
11rdc11
Quote:

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.
• December 26th 2009, 04:59 PM
HallsofIvy
Quote:

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.

Quote:

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.