1. ## complex plane

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2. Originally Posted by yellow4321
from previous question im told |a|=1
if F is some map F:complex-->complex defined by F(z)=az how can i show the map F is a rotation. i know it will be a rotation because the angles will be added but i get confused about cos(theta)+isin(theta) where abouts on the argand diagram would this lie if cos(theta)+isin(theta)=1
then suppose a=i what would be the rotation then?
If $|a|=1$ then $a=e^{i\theta}$ and so if $z = |z|e^{i\arg (z)}$ then $az = |z|e^{i\theta}e^{i\arg(z)} = |z|e^{i(\theta + \arg (z))}$ which is the same complex number by rotated $\theta$ .

This is Mine 85th Post!!!

3. Originally Posted by ThePerfectHacker
If $|a|=1$ then $a=e^{i\theta}$ and so if $z = |z|e^{i\arg (z)}$ then $az = |z|e^{i\theta}e^{i\arg(z)} = |z|e^{i(\theta + \arg (z))}$ which is the same complex number by rotated $\theta$ .
It goes without saying (well, it nearly did ) that there's NO rotation if a = 1.

4. Originally Posted by mr fantastic
It goes without saying (well, it nearly did ) that there's NO rotation if a = 1.
It is, it is called a "trivial rotation".

5. thank you what if a=i ?

6. Originally Posted by yellow4321
thank you what if a=i ?
Let us see what happens. Pick a number, say $1+0i$ and graph it on the complex plane. If you multiply this by $i$ we get $0+i$. Pick a point as $-1+0i$ multiply it by $i$ and it becomes $-i$. Try it for $\pm i$. Can you guess what type of rotation this is?