3 - 4i equals

\(\text { Let } \mathbf{z}=\mathbf{3}-\mathbf{4} \mathbf{i}\\ |\mathbf{z}|=\sqrt{3^{2}+4^{2}}=5\left\{|\mathbf{a}-\mathbf{b i}|=\sqrt{\mathbf{a}^{2}+\mathbf{b}^{2}}\right\}\\ \mathbf{z}=3-4 \mathbf{i}=\mathbf{r e}^{\mathrm{i} \theta}\\ \boldsymbol{\theta}=\mathbf{a r g} \& \mathbf{r}=|\mathbf{z}|\\ \theta=\tan ^{-1} \frac{\mathbf{b}}{\mathbf{a}}=\tan ^{-1}\left(\frac{-4}{3}\right)=\pi-\tan ^{-1}\left(\frac{4}{3}\right) \quad \ldots\left\{\tan ^{-1}(-\theta)=\pi-\tan ^{-1}(\theta)\right\}\\ \theta \text { lies in the fourth quadrant. }\\ \text { Argument }=-\mathbf{a}=-\left|\pi-\tan ^{-1}\left(\frac{4}{3}\right)\right|\\ \mathbf{z}=\mathbf{r e}^{\mathrm{i} \theta}\\ \mathbf{z}=5 \mathbf{e}^{-\mathbf{i}\left[\pi-\tan ^{-1}\left(\frac{4}{3}\right)\right]}\)

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