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Compex Numbers

Is my working correct? PART B

After drawing a diagram,

tan⁡θ=−3λ2+2λ−3λ+2 \tan \theta = \frac{-3\lambda^2 +2\lambda}{-3\lambda+2}

Arg[w]=tan⁡θ=π2 Arg[w] = \tan \theta = \frac{\pi}{2}

−3λ2+2λ−3λ+2=π2 \frac{-3\lambda^2 +2\lambda}{-3\lambda+2} = \frac{\pi}{2}

λ(−3λ+2)(−3λ+2)=π2 \frac{\lambda(-3\lambda +2)}{(-3\lambda +2)}= \frac{\pi}{2}

λ=π2 \lambda = \frac{\pi}{2}

552b6226b10fdb97b730432c04f73027.png
Reply 1
Avatar for Zacken
Zacken
22
Original post by Yunique
Is my working correct? PART B

After drawing a diagram


Nopes. You're overcomplicating this. arg⁡w=π2\arg{w} = \frac{\pi}{2} means the complex number w lies on the vertical imaginary axis. That means, it's real part is 0.

What's the real part of ww? (in terms of λ\lambda) set it equal to 0 and solve.
Reply 2
Avatar for Yunique
Yunique
OP
2
Original post by Zacken
Nopes. You're overcomplicating this. arg⁡w=π2\arg{w} = \frac{\pi}{2} means the complex number w lies on the vertical imaginary axis. That means, it's real part is 0.

What's the real part of ww? (in terms of λ\lambda) set it equal to 0 and solve.


True, but surely there is a way to soIve it using the method I was trying to take?
Reply 3
Avatar for Zacken
Zacken
22
Original post by Yunique
True, but surely there is a way to soIve it using the method I was trying to take?


It'll boil down to the same thing. If the argument is π2\frac{\pi}{2} then your argument is arg⁡(w)=arctan⁡ℑ(w)ℜ(z)\arg(w) = \arctan \frac{\Im(w)}{\Re(z)}. Apply the tangent to both sides: tan⁡arg⁡(w)=ℑ(w)ℜ(w)\tan \arg(w) = \frac{\Im(w)}{\Re(w)}.

tan⁡(π2)\tan(\frac{\pi}{2}) is undefined. So we need the denominator to be 0, i.e: ℜ(w)=0\Re(w) = 0.

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