# Math Help - Finding r and the exact value

1. ## 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.

2. There are formulae for these
So if your complex number is a+bi,

r=√(a²+b²)

Now for θ you need to make sure you get your answer in the right quadrant.
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.

3. 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?

4. 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)