# Complex Numbers school work

1. Sep 17, 2007

1. The problem statement, all variables and given/known data

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

2. Relevant equations
z=a+bi
$$arg z=arg tan \frac {b}{a}$$

3. 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?

Last edited: Sep 17, 2007
2. Sep 17, 2007

### Hurkyl

Staff Emeritus
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?)

3. Sep 17, 2007

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 doesnt work.
So my final answer, b=1.732

Last edited: Sep 17, 2007
4. Sep 17, 2007

### HallsofIvy

Staff Emeritus
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."

5. Sep 18, 2007