Help with complex number derivation

[diracy]

Homework Statement

(a) Suppose the segment connecting (a,b) to (0,0) has length r[itex]_{1}[/itex] and forms an angle [itex]\theta[/itex][itex]_{1}[/itex] with the positive side of the x-axis. Suppose the segment connecting (c,d) to (0,0) has a length r[itex]_{2}[/itex] and forms an angle [itex]\theta[/itex][itex]_{2}[/itex] with the positive side of the x-axis. Now let (a+bi)(c+di)=x+yi. Show that the length of the segment connecting (x,y) to the origin is r[itex]_{1}[/itex]r[itex]_{2}[/itex] and the angle formed is [itex]\theta[/itex][itex]_{1}[/itex]+[itex]\theta[/itex][itex]_{2}[/itex].

(b) Use the result from (a) to find a complex number z[itex]\in[/itex]C such that z^2=i.

Homework Equations

The Attempt at a Solution

(a+bi)(c+di)=x+yi
ac+adi+bci+bd(i[itex]^{2}[/itex])=x+yi
ac+adi+bci-bd=x+yi
(ac-bd)+(ad+bc)=x+yi
x=(ac-bd), y=(ad+bc)

I'm not sure where to go from here. Just looking for some help. Thanks!

 

[eumyang]

Remember the formulas for going from standard form of a complex number to trig form. If z = a + bi = r(cos θ + i sin θ), then
[itex]r = \sqrt{a^2 + b^2}[/itex],
a = r cos θ and b = r sin θ, and
[itex]\tan \theta = \frac{b}{a}[/itex].

If x=(ac-bd) and y=(ad+bc), then what is
[itex]\sqrt{x^2 + y^2}[/itex]? You'll need to make it equal to r₁r₂.

For the angle, you'll need the tangent of a sum identity.
[itex]\tan (\theta_1 + \theta_2) = \frac{\tan \theta_1 + \tan \theta_2}{1 - \tan \theta_1 \tan \theta_2}[/itex]

 

[diracy]

Remember the formulas for going from standard form of a complex number to trig form. If z = a + bi = r(cos θ + i sin θ), then
[itex]r = \sqrt{a^2 + b^2}[/itex],
a = r cos θ and b = r sin θ, and
[itex]\tan \theta = \frac{b}{a}[/itex].

If x=(ac-bd) and y=(ad+bc), then what is
[itex]\sqrt{x^2 + y^2}[/itex]? You'll need to make it equal to r₁r₂.

For the angle, you'll need the tangent of a sum identity.
[itex]\tan (\theta_1 + \theta_2) = \frac{\tan \theta_1 + \tan \theta_2}{1 - \tan \theta_1 \tan \theta_2}[/itex]

I'm honestly not getting very far with this. Could you help me out a little more?

 

[diracy]

Ok, I got the first part. Now I need to prove the angle part. Any help?

 

[eumyang]

If x=(ac-bd) and y=(ad+bc), then what is
[itex]\sqrt{x^2 + y^2}[/itex]?

Start with this. Plug in what x and y equals into the square root and expand the radicand. Show us what you get.

EDIT: Never mind. You said that you got this part.

For the angle, you'll need the tangent of a sum identity.
[itex]\tan (\theta_1 + \theta_2) = \frac{\tan \theta_1 + \tan \theta_2}{1 - \tan \theta_1 \tan \theta_2}[/itex]

[itex]\tan \theta_1 = \frac{b}{a}[/itex]
and
[itex]\tan \theta_2 = \frac{d}{c}[/itex]
Plug these into the formula above and simplify. Somehow you'll have to simplify to y/x. (Remember that you had x and y can be expressed in terms of a, b, c, and d.)

 

[diracy]

Start with this. Plug in what x and y equals into the square root and expand the radicand. Show us what you get.

EDIT: Never mind. You said that you got this part.



[itex]\tan \theta_1 = \frac{b}{a}[/itex]
and
[itex]\tan \theta_2 = \frac{d}{c}[/itex]
Plug these into the formula above and simplify. Somehow you'll have to simplify to y/x. (Remember that you had x and y can be expressed in terms of a, b, c, and d.)

Hmmm...

After I plug those in, what methods can I use to simply that expression?

 

[eumyang]

After plugging those in, try multiplying the numerator and denominator of this complex fraction by ac.

(Going to bed now... :zzz:)