If z=r(cos θ + isin θ), where r>0 and 0< θ < 1/2π, find in terms of r and θ the modulus and principal argument of....

a) -z

I started off by: -z=-r(cos θ + isin θ) ---> = r(-cos θ - isin θ) = r(-cos(-θ) + isin (-θ)).

a= -rcos -θ b= rsin -θ

√(-rcos(-θ))^2 + (rsin (-θ))^2) = r

I have no idea how to continue from this. I know the formula for the argument is --> tan θ = b/a

Plugging in the values I have would result in tan θ = -tan -θ

The answer is supposed to be Mod= r Arg= θ - π


$$-z = z(-1) = re^{i\theta}e^{i\pi} = re^{i(\theta+\pi)}$$ Hence, Modulus is $r$

Now for principal argument, the angle by definition must lie in $(-\pi, \pi]$. Therefore, we need to translate $\pi + \theta $ to this interval. Since $0 \lt \theta \lt \pi/2$, we use $$\tan(\theta) = \tan(\theta - 2\pi)$$ Giving us our primary argument as $\theta - \pi$

  • $\begingroup$ How does -1 = 𝑒𝑖𝜋 ?? $\endgroup$ – Aino Apr 22 at 15:01
  • $\begingroup$ $e^{i\pi} = \cos{\pi} + i\sin{\pi} = -1 + 0 = -1$ $\endgroup$ – Dhanvi Sreenivasan Apr 23 at 4:45

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