Finding Polar Form Expressions: -3-3i & 2√3-2i

[ver_mathstats]

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)+isin(-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)+isin(-π/6)]. Here I thought if we get a negative we must add π to it.

Thank you.

 

[WWGD]

Notice you need to take into consideration the quadrants where your x,y coordinate are in order to find the angles. When you do $tan^{-1}$ the signs cancel out so that you "lose information".

 

[Ray Vickson]

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)+isin(-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)+isin(-π/6)]. Here I thought if we get a negative we must add π to it.

Thank you.

In problem (a) your $\theta = \pi/4$ is incorrect, because it is in the wrong quadrant. Your $z = -3 - 3i$ has negative real and imaginary parts, so must be in the third quadrant. So, either you allow yourself to use negative angles, in which case you need $-\pi/2 > \theta > -\pi$ (that is $\theta$ lies between $-180^o$ and $-90^o$) or else you restrict yourself to positive angle, in which case $\pi < \theta < 3\pi/2$ (that is, $\theta$ lies between $180^o$ and $270^o$). You need the solution of the equation $\tan \theta = 1$ that lies in the third quadrant.

 

[RPinPA]

My two cents:

For any given point in the complex plane $x + iy$, then $\tan(\theta) = y/x$.

So notice that the same value of tangent is obtained from $(-y)/(-x)$. That point $-x-iy$ is 180 degrees away from $x + iy$. Given any particular value for the tangent, there are two points 180 degrees apart that have that same tangent.

The standard arctan function is defined to return a value between $-\pi/2$ and $\pi/2$. Those angles correspond to a positive $x$. So the easiest thing to do is check the sign of $x$. If it's negative, then you need the other angle with that same tangent, which you obtain by adding or subtracting $\pi$ radians.

Applying that rule to your problem, you see $x = -3$, so the correct angle is $\pi/4 + \pi = 5\pi/4$ or equivalently $\pi/4 - \pi = -3\pi/4$.

Again, to summarize what's going on here: there are two angles 180 degrees apart that have the same tangent, and the standard arctan function returns only one of them. You have to check if you want the other one.