Question #191f5

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Answer 1 · 2017-04-06T20:21:08

$- 4 + 5 i \to \sqrt{41} (cos ({128.7}^{\circ}) + i sin ({128.7}^{\circ}))$ $-4 + 5i to sqrt41(cos(128.7^@) + isin(128.7^@))$

To convert $a + b i \to r (cos (θ) + i sin (θ))$ $a + bi to r(cos(theta)+isin(theta))$

$r = \sqrt{a^{2} + b^{2}}$ $r = sqrt(a^2 + b^2)$

$r = \sqrt{- 4^{2} + 5^{2}}$ $r = sqrt(-4^2+5^2)$

$r = \sqrt{16 + 25}$ $r = sqrt(16+25)$

$r = \sqrt{41}$ $r = sqrt(41)$

Because "a" is negative and "b" is positive we compensate for the second quadrant by adding $180^{\circ}$ $180^@$ :

$θ = {tan}^{- 1} (\frac{b}{a}) + 180^{\circ}$ $theta = tan^-1(b/a)+180^@$

$θ = {tan}^{- 1} (\frac{5}{- 4}) + 180^{\circ}$ $theta = tan^-1(5/-4)+180^@$

$θ \approx {128.7}^{\circ}$ $theta ~~ 128.7^@$

$- 4 + 5 i \to \sqrt{41} (cos ({128.7}^{\circ}) + i sin ({128.87}^{\circ}))$ $-4 + 5i to sqrt41(cos(128.7^@) + isin(128.87^@))$