Finding r and the exact value

• Apr 20th 2009, 09:03 AM
goldenroll
Finding r and the exact value
I have no idea how to do these problems on my review, can someone please explain how to do these.

1. -2+5i= r(cosθ+isinθ), find θ accurate to 3 decimal places. Be sure your answer is in the correct quadrant.

2. If θis an angle in the third quadrant and sinθ= -1/5 find the exact value of tan
2θ. I did not rationalize the denominator.

3. If 2+5i =
r(cosθ+isinθ), find r.

• Apr 20th 2009, 12:13 PM
charlie
There are formulae for these :)
So if your complex number is a+bi,

r=√(aČ+bČ)

So draw your complex number on an argand diagram and see where it is.

If it's in the 1st quadrant, eg. 2+5i, then θ = arctan(b/a)
2nd quadrant, eg. -2+5i, then θ = pi - arctan(b/a)
3rd quadrant, eg. -2-5i, then θ = arctan(b/a) - pi
4th quadrant, eg. 2-5i, then θ = - arctan(b/a)

I hope this helps.
• Apr 20th 2009, 12:47 PM
goldenroll
Ok, so if I got you correct then

3. square root(29)

1.
Also for this one I looked up some stuff online but I think I got it wrong

tan^(-1)(y/x)=5/2=68.2(rounded)
180 -68=112
Square root 29(cos(thata)+isin(thata)=5.39
5.39(cos(112)+isin(112)=-2?
• Apr 20th 2009, 01:04 PM
charlie
Quote:

Originally Posted by goldenroll
Ok, so if I got you correct then

3. square root(29)

1.
Also for this one I looked up some stuff online but I think I got it wrong

tan^(-1)(y/x)=5/2=68.2(rounded)
180 -68=112
Square root 29(cos(thata)+isin(thata)=5.39
5.39(cos(112)+isin(112)=-2?

Yup, Q3. is right :)

For Q1. θ is usually calculated in radians, so you'll have to put your calculator into radians. You've got the right idea though.
(I forgot to mention that when you're working out arctan(b/a), always take the positive of b and a. What you do with the quadrant thing will make sure the answer's right. You did that anyways though :) )

So -2+5i is in the 2nd quadrant, so pi - arctan(5/2) = 1.951 to 3dp.

You can check this: √29(cos(1.951)+ i sin(1.951)) = -1.998 + 5.001i
(It's not exact because your theta was rounded)