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What’s the complex number in polar form with argument theta between 0 and 2pie?
8+8i

What’s the complex number in polar form with argument theta between 0 and 2pie? 8+8i-example-1
User Xuan
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1 Answer

7 votes

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Answer:

45. (8√2)∠π/4

46. (7√2)∠3π/4

Explanation:

The polar form of a+bi is ...

a +bi ⇔ (√(a²+b²))∠arctan(b/a)

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45. 8 +8i = (√(8² +8²))∠arctan(8/8) = (8√2)∠π/4

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46. -7 +7i = (√((-7)² +7²))∠arctan(7/-7) = (7√2)∠3π/4

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Additional comment

The arctan function usually returns a value in the range -π/2 to π/2. The quadrant of the desired result can be determined by the signs of the components of the vector. For negative real parts, π needs to be added to the value that is usually returned by the arctan function.

User Evaristokbza
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