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RoseWind [281]
3 years ago
6

Can someone please help me with this question?

Mathematics
1 answer:
Genrish500 [490]3 years ago
4 0
It’s numberrrrr 333333333333333
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8

Step-by-step explanation:

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3 years ago
Convert,the complex number into polar form: 4+4i
kow [346]
Z = a + bi
z = 4 + 4i

r² = a² + b²
r² = (4)² + (4)²
r² = 16 + 16
r² = 32
 r = 4√(2)
 r = 4(1.414)
 r = 5.656

cos\theta = \frac{a}{r}
cos\theta = \frac{4}{4\sqrt{2}}
cos\theta = \frac{4}{4\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}}
cos\theta = \frac{4\sqrt{2}}{4\sqrt{4}}
cos\theta = \frac{4\sqrt{2}}{4(2)}
cos\theta = \frac{4\sqrt{2}}{8}
cos\theta = \frac{\sqrt{2}}{2}
2(cos\theta) = 2(\frac{\sqrt{2}}{2})
2cos\theta = \sqrt{2}
2cos\theta = 1.414

sin\theta = \frac{b}{r}
sin\theta = \frac{4}{4\sqrt{2}}
sin\theta = \frac{4}{4\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}}
sin\theta = \frac{4\sqrt{2}}{4\sqrt{4}}
sin\theta = \frac{4\sqrt{2}}{4(2)}
sin\theta = \frac{4\sqrt{2}}{8}
sin\theta = \frac{\sqrt{2}}{2}
2(sin\theta) = 2(\frac{\sqrt{2}}{2})
2sin\theta = \sqrt{2}
2sin\theta = 1.414

z = a + bi
z = rcosθ + (rsinθ)i
z = r(cosθ + i sinθ)

z = 4 + 4i
z = 5.656cosθ + (5.656sinθ)i
z = 5.656(cosθ + i sinθ)
z = 5.656(cos45 + i sin45)

\theta = tan^{-1}\frac{b}{a}
\theta = tan^{-1}\frac{4}{4}
\theta = tan^{-1}(1)
\theta = 45

The polar form of 4 + 4i is approximately equal to 5.656(cos45 + i sin45).
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3 years ago
What is a binomial polynomial?
Gemiola [76]

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