Finding complex number from polar coordinates

**shadow_2145** · August 28th 2008, 11:40 PM

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

**Moo** · August 28th 2008, 11:42 PM

Hi !

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

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.

**shadow_2145** · August 28th 2008, 11:51 PM

Is the answer to (a) $\sqrt{2} + \sqrt{2}i$ ?

**shadow_2145** · August 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

**wingless** · August 29th 2008, 12:12 AM

How did you find $\sqrt{2} + \sqrt{2}i$ ?

$r\cdot (\cos \theta + i \sin \theta ) = \sqrt{3}\cdot (\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}) = \frac{\sqrt{6}}{2}+\frac{\sqrt{6}}{2}i$

**shadow_2145** · August 29th 2008, 12:23 AM

Very sleepy and dumb mistake. I see now, thanks.

**Prove It** · August 29th 2008, 04:12 AM

Originally Posted by shadow_2145

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

For the second, you'd get $z=4(\cos(\frac{-\pi}{2})+i \sin(\frac{-\pi}{2}))=4(0-i)=-4i$

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