# Finding complex number from polar coordinates

• Aug 28th 2008, 11:40 PM
Finding complex number from polar coordinates
Thanks for all the help on the previous problems, I figured out all my problems! Now, I'm stuck on another kinda simple area. Converting polar coordinates to complex numbers...

Give the complex number whose polar coordinates $(r,\theta)$ are

(a) $(\sqrt{3},\frac{\pi}{4})$

(b) $(4, \frac{-\pi}{2})$

If someone could just walk me through one of the problems, I would greatly appreciate it! I have 6 of these to do, and I'm pretty sure I can figure them all out if someone can provide me with a walk-through on one! Thanks, and good night! zzzz
• Aug 28th 2008, 11:42 PM
Moo
Hi !

It's just $z=r \cdot e^{i \theta}=r \cdot (\cos(\theta)+i \sin(\theta))$

(Tongueout)

Why ?
Because r represents the distance from the origin to the point, which corresponds to the modulus you know $\sqrt{x^2+y^2}$. Make a sketch, you will understand ! :)
For the angles, it's the same, look at a sketch.
• Aug 28th 2008, 11:51 PM
Is the answer to (a) $\sqrt{2} + \sqrt{2}i$ ?
• Aug 28th 2008, 11:53 PM
Thanks Moo, I actually used that to figure out my answer before looking at your post! I think it is correct... Q_Q
• Aug 29th 2008, 12:12 AM
wingless
How did you find $\sqrt{2} + \sqrt{2}i$ ?

$r\cdot (\cos \theta + i \sin \theta ) = \sqrt{3}\cdot (\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}) = \frac{\sqrt{6}}{2}+\frac{\sqrt{6}}{2}i$
• Aug 29th 2008, 12:23 AM
Very sleepy and dumb mistake. I see now, thanks.
• Aug 29th 2008, 04:12 AM
Prove It
Quote:

Give the complex number whose polar coordinates $(r,\theta)$ are
(a) $(\sqrt{3},\frac{\pi}{4})$
(b) $(4, \frac{-\pi}{2})$
For the second, you'd get $z=4(\cos(\frac{-\pi}{2})+i \sin(\frac{-\pi}{2}))=4(0-i)=-4i$