# The principal argument of a complex number in polar form

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 $$\theta - \pi$$
• $e^{i\pi} = \cos{\pi} + i\sin{\pi} = -1 + 0 = -1$ – Dhanvi Sreenivasan Apr 23 at 4:45