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this is a question on my homework that I am just lost with. Some direction would be greatly appreciated.

Write $z_1$ and $z_2$ in polar form, and then find the product $z_{1}z_{2}$ and the quotients $z_1/z_2$ and $1/z_1$. (Express your answers in polar form.)
$$z_1 = \sqrt{2}-\sqrt{2}i, \ z_2 = 3-3i$$

I have calculated the modulus for each as $2$ and $3\sqrt{2}$ respectively. This I believe is correct as it is a simple formula. I then calculated the theta of each as $\pi/4$ for both of them, this was using $\arctan(\sqrt{2}/\sqrt{2})$ and $\arctan(3/3)$ respectively. Now when I drop them into place for $z_{1}z_{2}$ I receive:

$$\begin{align*} 6 \sqrt{2}\Bigl(\cos(\pi/2)+i \sin(\pi/2)\Bigr)&= 6\sqrt{2}\Bigl(0+i1\Bigr)\\ &= 6\sqrt{2}i\\ &= z_{1}z_{2} \end{align*}$$ Am I just making incorrect calculations or missing something? Apologies for my lack of MathJaX knowledge. Thanks for any help.

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up vote 3 down vote accepted

You are doing the wrong arctangents. Since $z_1 = \sqrt{2} - \sqrt{2}i$, then $z_i$ is in the "fourth" quadrant. (Also, after you get the arctangent, you need to make sure it is in the correct range (which in this case would be between $-\pi/2$ and $\pi/2$).

You should be computing $$\arctan\left(-\frac{\sqrt{2}}{\sqrt{2}}\right) = \arctan(-1) = -\frac{\pi}{4}.$$ Likewise for $z_2$: the argument for $z_2$ is not $\arctan(3/3)$, but $\arctan(-3/3) = \arctan(-1) = -\frac{\pi}{4}$.

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That makes sense, I can't believe I missed the sign. So then it would be correct to have $$-6\sqrt{2}i = z_{1}z_{2}$$? Thanks for your help. – Mike Soule Apr 11 '11 at 22:38
@Mike Soule: Yeah, looks right for the product. – Arturo Magidin Apr 12 '11 at 1:58

$z_1=2e^{-\pi i/4}, z_2=3\sqrt{2}e^{-\pi i/4}$. $z_1z_2=6\sqrt{2}e^{-\pi i/2}=-6\sqrt{2}i$

your mistake is the angle, which is $-\pi/4$ (an arctangent of -1 which is in the fourth quadrant, as both of your points have positive real part and negative imaginary part. draw a picture of the points in the plane)

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Using the $\arctan$, you need to account for the fact that $\tan(\theta)=\tan(\theta+\pi)$ so you need to resolve the ambiguity. – Ross Millikan Apr 11 '11 at 22:08
no you dont; they are both geometrically in the 4th quadrant – yoyo Apr 12 '11 at 2:06
Both OP's points are, but $\frac{3\pi }{4}$ has the same tangent. If one of the points had been $-\sqrt{2}+\sqrt{2}i$ the calculation would have been the same. – Ross Millikan Apr 12 '11 at 2:11

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