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Express -3-3i in polar form.

I know that r=3√2.

And I understand that now we take tan^-1(b/a) which I did. tan^-1(-3/-3) = π/4. So I put my answer as z = 3√2 [cos(π/4) + isin(π/4)].

However the answer manual told me this was incorrect I am unsure of where I went wrong?

3√2[cos(-3π/4)+

Then the same thing happened again for express 2√3-2i in polar form.

I know that r=4.

And I understand that now we take tan^-1(b/a) which I did. tan^-1(-2/(2√3)) = -π/6. So I put my answer as z = 4 [cos(5π/6) + isin(5π/6)].

But the answer was 4[cos(-π/6)+

Thank you.

I know that r=3√2.

And I understand that now we take tan^-1(b/a) which I did. tan^-1(-3/-3) = π/4. So I put my answer as z = 3√2 [cos(π/4) + isin(π/4)].

However the answer manual told me this was incorrect I am unsure of where I went wrong?

3√2[cos(-3π/4)+

*i*sin(-3π/4)] this was the answer in the manual.Then the same thing happened again for express 2√3-2i in polar form.

I know that r=4.

And I understand that now we take tan^-1(b/a) which I did. tan^-1(-2/(2√3)) = -π/6. So I put my answer as z = 4 [cos(5π/6) + isin(5π/6)].

But the answer was 4[cos(-π/6)+

*i*sin(-π/6)]. Here I thought if we get a negative we must add π to it.Thank you.

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