Sketching Regions in the Complex Plane | Complex Numbers

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In this video, we look at sketching regions in the complex plane. My students found problems of this type quite difficult, so hopefully, this video helps.
Practice Problems: theorembmath.f...
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Опубликовано:

23 сен 2024

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Комментарии : 46

@KyleBroder 3 года назад

👍 If you like the video, hit the like button and subscribe to the channel if you want to show your support/want to learn some mathematics.

@NikitaMelik-Marutov 9 месяцев назад

you have a very calm voice. It really helps in studing process. Thank you

@pelasgeuspelasgeus4634 9 месяцев назад

Can you please explain why in the complex plane the imaginary axis unit is depicted to have same length as the real axis unit 1? Is it an arbitrary convention or is it something else?

@amabonge1442 18 дней назад

Hi Kyle thank you for the video, it was really helpful... Please assist, where can I find the memo for the practice problems?

@ProfLearn 8 месяцев назад

HI SIR WHICH SOFTWARE ARE U USING WHEN WRITTING

@leifsy2142 Год назад

In the last item, is the border included in the shaded region?

@jasma987 7 месяцев назад

@alanwatts1746 4 года назад

Example 3 was really helpful, I was really struggling with these from lectures.

@KyleBroder 4 года назад

Glad you found it useful!

@ArtsandGadgetsStuff 5 месяцев назад

Example 3 and 4 could be done in a very easy way... Let me explain it with example 4, 1. Plot points 3i and 1 2. Choose equality if the given condition |z-3i| > |z-1|, will be |z-3i| = |z-1| Now, if you observe this, we can notice that it is perpendicular bisector equation, which is equidistant from 3i and 1. 3. Draw the perpendicular bisector. You can get its slope as it is perpendicular to line joining 3i and 1. (To get slope, first find the slope of 3i and 1. Let those points be (0,3) and (1,0), slope is -3, then slope of line is 1/3) 4. Remove equality Now, observing |z-3i| > |z-1|, the distance of locus from 3i more compared to 1, then, the locus will be the area covering the side of the biscetor facing (1,0) And, that's how we do it. It might seem difficult at first, but once you try it, it's very simple. It's EXTREMELY useful if you just want a rough sketch.

@GeorgeNotTheMonkey 2 года назад

In example 4 isn't the boundary supposed to be a dotted line?

@KyleBroder 2 года назад

Let me check :)

@gavincourts854 4 года назад

How would you describe the complement of an annulus?

@KyleBroder 3 года назад

Here is one way: If the annulus you want the complement of is A = \{ z \in \mathbb{C} : r < | z | < R \}. Then the complement is \{ z \in \mathbb{C} : | z | \leq r \} \cup \{ z \in \mathbb{C} : | z | \geq R \}. (I hope you can read the above LaTeX).

@gavincourts854 4 года назад

The tute notes are amazing by the way, you should continue with them.