How do you divide $\frac{7 - 9 i}{6 + i}$ $(7-9i)/(6+i)$ in trigonometric form?

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Answer 1 · 2018-05-03T20:53:56

$= \frac{33}{37} - \frac{61}{37} i$ $= 33/37-61/37i$

$\frac{7 - 9 i}{6 + i} ∣ \cdot (6 - i)$ $(7-9i)/(6+i) | *(6-i)$

$\frac{(7 - 9 i) (6 - i)}{(6 + i) (6 - i)}$ $((7-9i)(6-i))/((6+i)(6-i))$

$\frac{42 - 61 i + 9 i^{2}}{36 - 6 i + 6 i - i^{2}}$ $(42-61i+9i^2)/(36-6i+6i-i^2)$

$\frac{42 - 61 i + 9 i^{2}}{36 - i^{2}}$ $(42-61i+9i^2)/(36-i^2)$

$\frac{42 - 9 - 61 i}{36 + 1}$ $(42-9-61i)/(36+1)$

$\frac{33 - 61 i}{37}$ $(33-61i)/(37)$

$= \frac{33}{37} - \frac{61}{37} i$ $= 33/37-61/37i$

How do you divide 7−9i6+i (7-9i)/(6+i) in trigonometric form?