Polar form (complex numbers)

[TsAmE]

Write -3 + 3i in polar form with argument between 0 and 2?.

Attempt:

-3 + 3i is in the 2nd quadrant so I got an angle of \(\displaystyle (\pi - \frac{\pi}{4}) = \frac{3\pi}{4}\)

\(\displaystyle = 3\sqrt{2}(cos(\frac{-3\pi}{4}) + isin(\frac{3\pi}{4}))\)

but the correct answer was:

\(\displaystyle 3\sqrt{2}(cos(\frac{3\pi}{4}) + isin(\frac{3\pi}{4}))\)

I dont get why the angle of the real part \(\displaystyle (cos(\frac{3\pi}{4}))\) is positive, since cos is negative in the 2nd quadrant?

 

[pka]

Write -3 + 3i in polar form with argument between 0 and 2?.
\(\displaystyle 3\sqrt{2}(cos(\frac{-3\pi}{4}) + isin(\frac{3\pi}{4}))\)
but the correct answer was:
\(\displaystyle 3\sqrt{2}(cos(\frac{3\pi}{4}) + isin(\frac{3\pi}{4}))\)

\(\displaystyle 3\sqrt{2}(cos(\frac{-3\pi}{4}) + isin(\frac{3\pi}{4}))=3\sqrt{2}(cos(\frac{3\pi}{4}) + isin(\frac{3\pi}{4}))\)

Those two are equal. You see \(\displaystyle \cos(t)=\cos(-t)\) it is an even function.

However, the second is the preferred form.

 

[TsAmE]

Those two are equal. You see \(\displaystyle \cos(t)=\cos(-t)\) it is an even function.

Isnt that the case only if t is in the 4th quadrant, and isnt \(\displaystyle cos\frac{-3\pi}{4}\) in the 3rd quadrant? (where cos is negative)

 

[Deleted member 4993]

3?/4 falls in the second quadrant ? cosine is negative.

(-3?/4) falls in the third quadrant ? cosine is negative. ? We get the same answer as above.

 

[TsAmE]

3?/4 falls in the second quadrant ? cosine is negative.

(-3?/4) falls in the third quadrant ? cosine is negative. ? We get the same answer as above.

I understand this but then in the final answer why is the cosine + and not - ? Im still confused

 

[Mrspi]

3?/4 falls in the second quadrant ? cosine is negative.

(-3?/4) falls in the third quadrant ? cosine is negative. ? We get the same answer as above.

I understand this but then in the final answer why is the cosine + and not - ? Im still confused

Write -3 + 3i in polar form with argument between 0 and 2?.

(3 pi)/4 is between 0 and 2 pi.

-(3 pi)/4 is NOT between 0 and 2 pi.

Does that help?

 

[TsAmE]

(3 pi)/4 is between 0 and 2 pi.

-(3 pi)/4 is NOT between 0 and 2 pi.

Does that help?

So basically \(\displaystyle cos(\frac{3\pi}{4}) = -cos(\frac{3\pi}{4})\), but you choose the positive one?

 

[Deleted member 4993]

(3 pi)/4 is between 0 and 2 pi.

-(3 pi)/4 is NOT between 0 and 2 pi.

Does that help?

So basically \(\displaystyle cos(\frac{3\pi}{4}) = -cos(\frac{3\pi}{4})\), but you choose the positive one? ..... No

\(\displaystyle cos(\frac{3\pi}{4}) = cos(-\frac{3\pi}{4})\),

 

[Deleted member 4993]

3?/4 falls in the second quadrant ? cosine is negative.

(-3?/4) falls in the third quadrant ? cosine is negative. ? We get the same answer as above.

I understand this but then in the final answer why is the cosine + and not - ? Im still confused

According to YOUR work:


cos(-3?/4) = -1/?2

The book saying

cos(3?/4) = -1/?2 .......... and both are correct

 

[Mrspi]

MANY times, there are multiple (often infinitely many) correct answers for a trig equation.

The problem often stipulates a restriction on what answers are acceptable.

Your problem did that.

The ARGUMENT (the angle you're using in the answer) is supposed to be between 0 and 2 pi.

That means you cannot use -(3pi/4) even though cos -(3pi/4) is the same as cos (3pi/4).....

One of those dumb rules....

Does not pay to fight it.