How do you write $2 - 3 i$ $2 - 3i$ in trigonometric form?

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How do you write $2 - 3 i$ $2 - 3i$ in trigonometric form?

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Answer 1 · 2016-01-27T14:46:00

$2 - 3 i = \sqrt{13} (cos θ + i sin θ)$ $2-3i = sqrt(13)(cos theta +isin theta)$

Explanation:

See Trigonometric form for a demonstration of how to do this.

In order to write a complex number $2 - 3 i$ $2 - 3i$ in trigonometric form we need to find the modulus $z$ $z$ , where $z = a + b i = | z | (cos θ + i sin θ)$ $z = a +bi = |z|(cos theta + i sin theta)$

$z = \sqrt{a^{2} + b^{2}} = \sqrt{2^{2} + {(- 3)}^{2}} = \sqrt{13}$ $z=sqrt(a^2+b^2) = sqrt(2^2 +(-3)^2) =sqrt(13)$

Then $2 - 3 i = \sqrt{13} (cos θ + i sin θ)$ $2-3i = sqrt(13)(cos theta +isin theta)$

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How do you write $2 - 3 i$ $2 - 3i$ in trigonometric form?

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How do you write $2 - 3 i$ $2 - 3i$ in trigonometric form?