Z_{1}=z_{2}

**dwsmith** · September 4th 2010, 05:02 PM

Suppose $z_1=r_1(\cos(\theta_1)+i\sin(\theta_1))$ and $z_2=r_2(\cos(\theta_2)+i\sin(\theta_2))$ . If $z_1=z_2$ , then how are $r_1 \ \mbox{and} \ r_2$ related? How are $\theta_1 \ \mbox{and} \ \theta_2$ related?

If $z_1=z_2$ , then $a_1=a_2 \ \mbox{and} \ b_1=b_2$ . Therefore, $r_1=r_2$ and $\theta_1=\theta_2$ since $a^2+b^2=r^2 \ \mbox{and} \ \theta=\tan\frac{b}{a}$ .

Is this correct?

**Ackbeet** · September 4th 2010, 05:06 PM

I don't think your answer takes into account fully the ambiguities inherent in polar coordinates. You have negative radii possibilities, as well as changing the angle by certain well-defined amounts. How does this change your picture?

**dwsmith** · September 4th 2010, 05:11 PM

But by definition, $z_1=z_2$ iff. $a_1=a_2 \ \mbox{and} \ b_1=b_2$ . Why can't I use that fact?

**Ackbeet** · September 4th 2010, 05:18 PM

Sure, but that's cartesian coordinates. As soon as you start talking about radii and angles (i.e., polar coordinates), the game is different. Example:

$-2\cos(\pi/4)=2\cos(5\pi/4).$

In fact, in polar coordinates, the point $-2\,\text{cis}\,\pi/4$ is the same point as $2\,\text{cis}\,5\pi/4.$

So from this, you can see that setting

$r_{1}\cos(\theta_{1})=r_{2}\cos(\theta_{2})$ is simply not going to imply that $r_{1}=r_{2}$ and $\theta_{1}=\theta_{2}.$

Even if you throw in the other equation (the y equation), you're not going to resolve all the ambiguities.

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