### Author Topic: 2020 TT1 Main Setting - Problem 1b  (Read 789 times)

#### A A

• Newbie
• • Posts: 3
• Karma: 0 ##### 2020 TT1 Main Setting - Problem 1b
« on: October 14, 2020, 05:42:27 PM »
Hello, can anyone explain the solution to Problem 1b: which of the complex roots are in the first complex quadrant?  Why is it all zn that are in the first complex quadrant?

Note (for context): Problem 1a: Find all the complex roots of the equation tanh(3z) = 1 + 2i

#### RunboZhang

• Sr. Member
•    • Posts: 51
• Karma: 0 ##### Re: 2020 TT1 Main Setting - Problem 1b
« Reply #1 on: October 14, 2020, 07:37:18 PM »
Firstly, for the first quadrant, we have $Re(z) > 0$ and $Im(z) > 0$.
Secondly, we have $z = \frac{1}{12}log(2) + (\frac{\pi}{8} + \frac{2\pi }{6} n)i$, $n \in \mathbb{Z}$ by part (a).
By combining the previous two conclusions we have, $Re(z) = \frac{1}{12}log(2) > 0$ since $log(2) > 0$. Also $Im(z) = (\frac{\pi}{8} + \frac{2\pi }{6} n) > 0$ when $n \ge 0$.
Therefore, as long as we have a non-negative $n$, our $z$ is in the first quadrant of complex plane.

#### A A

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• Karma: 0 ##### Re: 2020 TT1 Main Setting - Problem 1b
« Reply #2 on: October 14, 2020, 09:32:09 PM »
Thank you!!