Polar form of z = 1 - i

[rose stone]

$z = 1 - i$
$r = \sqrt{1^2 + (-1)^2} = \sqrt{2}$
$\phi = \arctan\left(\frac{-1}{1}\right) + 2\pi$
When I should calculate $\phi$ I don't know what is right. Some times you add $2\pi$ and sometimes not...
Is this right??

[toxic thicket]

Put the math parts in $$ to have it latex rendered

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$z = 1 - i$

$r = \sqrt{1^2 + (-1)^2} = \sqrt{2}$

$\phi = \arctan\left(\frac{-1}{1}\right) + 2\pi$

When I should calculate $\phi$? I don't know what is right. Some times you add $2\pi$ and sometimes not...
Is this right??

[smoky python]

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[rose stone]

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