Question 3
Let $z=-\sqrt{3}+i$. The value of $\operatorname{Arg}\left(z^{4}\right)$ is
That’s it. Step through the hints below to see why.
Not quite. You powered the modulus only and kept $\operatorname{Arg}(z)=\frac{5\pi}{6}$.
Not quite. You subtracted $2\pi$ once, $\frac{10\pi}{3}-2\pi=\frac{4\pi}{3}$, which is still outside the principal range.
Not quite. You didn’t reduce $4\times\frac{5\pi}{6}$ at all.
Use the calculator’s $\operatorname{Arg}$ function.
Set Complex Mode a+bi and Angle Rad. Then OPTN → Complex (F3) → Arg (F3), and enter $\operatorname{Arg}\left(\left(-\sqrt3+i\right)^4\right)$.
$$\operatorname{Arg}\left(\left(-\sqrt3+i\right)^4\right)=-\frac{2\pi}{3}$$
The calculator returns the principal argument directly, so no reducing is needed.
$$\operatorname{Arg}\left(\left(-\sqrt3+i\right)^4\right)=-\frac{2\pi}{3}$$
$\operatorname{Arg}\left(z^4\right)=-\frac{2\pi}{3}$.
Only mark this done when you could do it without help. Reading the solution does not count.
A Tangent Tuition mock exam question. It is not a past QCAA question, and Tangent Tuition is not affiliated with the QCAA.
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