How do you write the complex number in trigonometric form $5 - 12 i$ $5-12i$ ?

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How do you write the complex number in trigonometric form $5 - 12 i$ $5-12i$ ?

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Answer 1 · 2017-06-06T10:35:15

In trigonometric form : $13 (cos 292.62 + i sin 292.62)$ $13(cos292.62+isin292.62)$

Explanation:

$Z = a + i b$ $Z=a+ib$ . Modulus: $| Z | = \sqrt{a^{2} + b^{2}}$ $|Z|=sqrt (a^2+b^2)$ ; Argument: $θ = {tan}^{- 1} (\frac{b}{a})$ $theta=tan^-1(b/a)$ Trigonometrical form : $Z = | Z | (cos θ + i sin θ)$ $Z =|Z|(costheta+isintheta)$

$Z = 5 - 12 i$ $Z=5-12i$ . Modulus $| Z | = \sqrt{5^{2} + {(- 12)}^{2}} = \sqrt{25 + 144} = \sqrt{169} = 13$ $|Z|=sqrt(5^2+(-12)^2) =sqrt(25+144)=sqrt169=13$

Argument: $tan α = \frac{12}{5} = 2.4$ $tan alpha = 12/5= 2.4$ . Z lies on fourth quadrant, $alpha =tan^-1(2.4) = 67.38^0 :. theta = 360-67.38=292.62^0 :. Z=13(cos292.62+isin292.62)$

In trigonometric form expressed as $13(cos292.62+isin292.62)$ [Ans]

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How do you write the complex number in trigonometric form $5 - 12 i$ $5-12i$ ?

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How do you write the complex number in trigonometric form $5 - 12 i$ $5-12i$ ?