graphing-functionspolar coordinates
I forgot the rules of adding angles when it comes to argand diagrams.
In the first quadrant, you add 90 degrees to whatever angle you get, what about Q2 Q3 Q4 ?
This picture will explain what i mean :
notice how 18.43 was added with 180 .. can someone list the rules for the other quadrants as well ?
what if i have -3i -3j for example ?
Thanks
Linear Bézier curve is simply a line given by parametric equation $R(t) = A+t(AB)$ , A being initial point and B being final point.
For Quadratic Bézier curve, take a look at the following picture.
Let the point between $P_1$ and $P_0$ be $Q_1$ and $P_1$ and $P_2$ be $Q_2$. Let our path be traced by $Q_0$. Then from above figure.
$$ \frac{P_0Q_1}{P_0P_1} = \frac{P_1Q_2}{P_1P_2} = \frac{Q_1Q_0}{Q_1Q2} = t \text{ (say)} $$
$$Q_1 = P_0 + t(P_0P_1), Q_2 = P_1 + t(P_1P_2)$$
So we have
$$Q_0 = Q_1 + t(Q_1Q_2) = P_0 + t(P_0P_1) + t(P_1 + t(P_1P_2) - (P_0 + t(P_0P_1)))$$
Have a look at more elaborate article on Wikipedia.
If you simply interpret what it means to say that $\arg(z-1)<\arg(z-i)$ geometrically, then it shouldn't be too hard to visualize the region. In each of the regions below, the argument of the point ($z$) minus the number (either $1$ or $i$) simply the angle from the dashed line to the arrow. It's easy to see that this angle is smaller from the complex number $1$ in exactly the blue regions.
Note that I'm assuming a branch cut at $\pi$ in the argument function so that, in the horizontal strip, $\arg(z-1)$ is definitely negative while $\arg(z-i)$ is positive.
Best Answer
When you measure the argument of a complex number you go counterclockwise from the x-axis.
So to find the argument you take $\arctan\left(\frac{b}{a}\right)$. However the range of the $\arctan x$ function is only from $[\frac{-\pi}{2},\frac{\pi}{2}]$. That is to say, you only get values between $-90^{\circ}$ and $90^{\circ}$ for whatever value you put in $\arctan x$. That means you are completely missing the $2$nd and $3$rd quadrant!
To account for the $2$nd and $3$rd quadrant we must first remember that $\tan \theta$ has period $\pi$. With that we realize that while $\theta+\pi$ has a different value from $\theta$; $\tan(\theta+\pi)=\tan(\theta)$. In fact it turns out $\tan(\theta +k\pi)=\tan(\theta),k\in\mathbb{Z}$. That means there are infinitely many solutions to $\tan x =\theta$ of the form $x=\arctan(\theta)+k\pi,k\in\mathbb{Z}$. But since $2\pi$ is one whole round around a circle all a lot of the solutions overlap and we are left with $2$ unique answers: $$x=\arctan(\theta)$$ $$x=\arctan(\theta)+\pi$$ However, your calculator does not know which solution you want and so it always gives you the first one. To get the second solution you simply add $\pi$ (or $180^{\circ}$ if you like degrees).