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galina1969 [7]
1 year ago
5

What is the square root of -2i?

Mathematics
1 answer:
Studentka2010 [4]1 year ago
4 0

Answer:

1-i and -1+i

Step-by-step explanation:

We are to find the square roots of z=0-2i. First, convert from Cartesian to polar form:

r=\sqrt{a^2+b^2}\\r=\sqrt{0^2+(-2)^2}\\r=\sqrt{0+4}\\r=\sqrt{4}\\r=2

\theta=tan^{-1}(\frac{b}{a})\\ \theta=tan^{-1}(\frac{-2}{0})\\\theta=\frac{3\pi}{2}

z=2(\cos\frac{3\pi}{2}+i\sin\frac{3\pi}{2})

Next, use the formula \displaystyle \sqrt[n]{r}\biggr[\cis\biggr(\frac{\theta+2\pi k}{n}\biggr)\biggr] where \displaystyle k=0,1,2,...\:,n-1 to find the square roots:

<u>When k=1</u>

<u />\displaystyle \sqrt[2]{2}\biggr[cis\biggr(\frac{\frac{3\pi}{2}+2\pi(1)}{2}\biggr)\biggr]

\displaystyle \sqrt{2}\biggr[cis\biggr(\frac{3\pi}{4}+\pi\biggr)\biggr]

\sqrt{2}\biggr(cis\frac{7\pi}{4}\biggr)

\sqrt{2}(\cos\frac{7\pi}{4}+i\sin\frac{7\pi}{4})\\ \\\sqrt{2}(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i)\\ \\1-i

<u>When k=0</u>

<u />\displaystyle \sqrt[2]{2}\biggr[cis\biggr(\frac{\frac{3\pi}{2}+2\pi(0)}{2}\biggr)\biggr]

\sqrt{2}\biggr(cis\frac{3\pi}{4}\biggr)

\sqrt{2}(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4})\\ \\\sqrt{2}(-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)\\ \\-1+i

Thus, the square roots of -2i are 1-i and -1+i

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Damm [24]

The angles in degrees to radian is as follows:

-54 degrees = -3π / 10 radian

<h3>How to convert from degree to radian?</h3>

The measurement is in degrees. Let's convert it to radian with respect to π.

Therefore,

180 degrees = π radian

-54 degrees = ?

cross multiply

Hence,

angle in radian = -54 × π / 180

angle in radian = - 54π / 180

angle in radian = - 6π / 20

angle in radian = -3π / 10 radian

learn more on radian here: brainly.com/question/22212006

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Answer:

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Step-by-step explanation:

According to question

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3 years ago
I need help with this question
creativ13 [48]
The 2nd option from the top
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HELP
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It looks like the parabola are going down, so  the coefficient should be negative,
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then we see that the parabola has x -intercepts (1,0) and (5,0)
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x=1, so x-1=0
and 
x=5, so x-5=0,
so (x-1)(x-5) and negative first coefficient -2 at the same time. 
We can see only in the 
B. <span>f(x) = –2(x – 5)(x – 1)</span>
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