# complex plane

• April 4th 2011, 10:58 AM
complex plane
hi guys

x + yi = cos(theta) + i sin(theta)

I understand this formula quite quite well...

But I have problem to think/understand how come if I have cos(theta) + i sin(theta) given I can come up with the x+yi form

Think in this way:

If someone gave me this input

cos(theta) + i sin(theta) == cos(0.6) + i sin(0.8) % in rad - and ask me to find x+yi version of this...

E.g. they look for x + yi , this is equal with 3 + 4i

How come I can come to this kind of conclusion? ...
• April 4th 2011, 11:09 AM
Plato
Quote:

If someone gave me this input
cos(theta) + i sin(theta) == cos(0.6) + i sin(0.8) % in rad - and ask me to find x+yi version of this...
E.g. they look for x + yi , this is equal with 3 + 4i
How come I can come to this kind of conclusion? ...

$\cos(0.6) + \mathbf{i} \sin(0.8)$ is in the $x +y \mathbf{i}$ form.
You see $x=\cos(0.8)~\&~y=\sin(0.8)$.
• April 4th 2011, 11:20 AM
NO NO NO ... I am asking how to get x=3 y=4 ? ... that what you say is quite easy, right?
• April 4th 2011, 11:28 AM
Plato
Quote:

NO NO NO ... I am asking how to get x=3 y=4 ? ... that what you say is quite easy, right?

I have absolutely no idea what that means.

Given $3+4\mathbf{i}$ let $\theta = \arctan \left( {\frac{4}{3}} \right)$.

Then $3+4\mathbf{i}=5\cos(\theta)+5\mathbf{i}\sin(\theta )$.
Is that what it means?
• April 4th 2011, 11:34 AM

Based on maths we know this formula: x + yi = cos(theta) + i sin(theta)

Now think in this way, someone has given you the right-hand side e.g., cos(0.6) + i sin(0.8)

And is asking you to find the left-hand side, which in this case is 3 + 4i

But as you may guess, could be: 6 + 8i, 12 + 16i etc. as well

So my question is given the right-hand side can we find left-hand side? Since given left-hand side we can find right-hand side precisely!
• April 4th 2011, 11:42 AM
Plato
Quote:

Based on maths we know this formula: x + yi = cos(theta) + i sin(theta)
Now think in this way, someone has given you the right-hand side e.g., cos(0.6) + i sin(0.8)
And is asking you to find the left-hand side, which in this case is 3 + 4i
But as you may guess, could be: 6 + 8i, 12 + 16i etc. as well
So my question is given the right-hand side can we find left-hand side? Since given left-hand side we can find right-hand side precisely!

Your got to be joking! Are you not?

Post an exact problem.
• April 4th 2011, 11:51 AM
veileen
"Based on maths we know this formula: x + yi = cos(theta) + i sin(theta)" This is not correct, $z=x+iy=\sqrt{x^2+y^2}(\cos \theta+i\sin \theta)$ (trigonometrical form)

So $x=\sqrt{x^2+y^2}\cos \theta$ and $y=\sqrt{x^2+y^2}\sin \theta$. (I hope you know that $\sqrt{x^2+y^2}=|z|$)

For your example: (which is wrong/ not complete)
1. False. cos(0.6) + i sin(0.8)=cos(theta) + i sin(theta) Try to guess what is wrong here! ^^
2. $z=3 + 4i=x+iy (x=3, y=4)$
$|z|=\sqrt{3^2+4^2}=\sqrt{25}=5$
$\cos \theta = \frac{x}{|z|}=\frac{3}{5}$
$\sin \theta = \frac {y}{|z|}=\frac{4}{5}$

You have some serious problems with theory. Learn it! >.<
• April 4th 2011, 11:53 AM
$z=x+iy=\sqrt{x^2+y^2}(\cos \theta+i\sin \theta)$ (trigonometrical form)