What is the exact value of b when arg z = 60 degrees?

[matadorqk]

Homework Statement

Given that z=(b+i)^2 where b is real and positive, find the exact value of b when arg z = 60 degrees.

Homework Equations

z=a+bi
arg z=arg tan \frac {b}{a}

The Attempt at a Solution

so I expanded my z=(b+i)^{2} so its
z=b^{2}-1+2bi

On other terms (please note the b here equals 2b, as it is the imaginary part, not the actual b)
so tan^{-1}\frac {b}{a}=60

tan60=\frac {b}{a}

atan60=b

**Dont get confused,
a=b^{2}-1

b=2b

Therefore, (b^{2}-1)tan60=2b

Here is where I am sort of confused, what now?

 

[Hurkyl]

It's good practice not to use the same letter for different purposes. We have 26 letters, two cases, and several styles to use... and that's just with one alphabet!


Anyways, you have a polynomial equation in one variable, don't you? I don't understand why you're stuck. (And don't you know the exact value of the tangent of 60 degrees?)

 

[matadorqk]

It's good practice not to use the same letter for different purposes. We have 26 letters, two cases, and several styles to use... and that's just with one alphabet!Anyways, you have a polynomial equation in one variable, don't you? I don't understand why you're stuck. (And don't you know the exact value of the tangent of 60 degrees?)

Ack, your right, stupid me :P.
So, 1.73...b^2-2b-1.73..=0
b=1.732
b=-0.577

I plotted both in, -0.577 doesn't work.
So my final answer, b=1.732

 

[HallsofIvy]

Notice that Hurkyl said "don't you know the exact value of the tangent of 60 degrees?"

Is there a reason for using the approximate value 1.732, rather than the exact value \sqrt{3}/2?

Your problem did say "find the exact value of b."

 

[weatherhead]

I think you'll find the exact value of tan 60 is sqrt(3) not sqrt(3)/2

 

[HallsofIvy]

Thanks. Went with sin(60) instead of tan(60).