![\stackrel{a}{4√(7)}~~ - ~~\stackrel{b}{9}i ~~ \begin{cases} r=√(a^2 + b^2)\\\\ \theta =\tan^(-1)\left( (b)/(a) \right) \end{cases} \\\\[-0.35em] ~\dotfill\\\\ r=\sqrt{(4√(7))^2 + 9^2}\implies r=√(16(7)+81)\implies r=√(193) \\\\\\ \theta =\tan^(-1)\left( \cfrac{9}{4√(7)} \right)\implies \theta \approx 40.38^o \\\\[-0.35em] ~\dotfill\\\\ z~~ = ~~√(193)( ~~ \cos(319.62^o)~~ + ~~i\sin(319.62^o) ~~ )\qquad \textit{\LARGE IV Quadrant}](https://img.qammunity.org/2023/formulas/mathematics/college/7kvjkvni27nhmlxmjml3qu461mbi6yqvvz.png)
let's take a quick look at the (a , bi) pair, "a" is positive whilst "b" is negative, that means the IV Quadrant, now let's keep in mind that tan⁻¹(x) has a range that lies on the I and IV Quadrants, so, if you check in your calculator for tan⁻¹(x) for those values, you'll notice it gives the one in the I Quadrant, namely 40.38°, however our coordinate pair is on the IV Quadrant, so we need its twin.
Make sure your calculator is in Degree mode.