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ra1l [238]
3 years ago
8

If

"absmiddle" class="latex-formula"> and z^{6}=a+bi, then a = _ and b = _
Mathematics
1 answer:
Luden [163]3 years ago
3 0

Rewriting z in polar form makes this trivial.

z=|z|e^{i\mathrm{arg}(z)}

We have

|z|=\sqrt{(-1)^2+(-\sqrt3)^2}=2

\mathrm{arg}(z)=\tan^{-1}(-1,-\sqrt3)=-\dfrac{2\pi}3

(not to be confused with the standard inverse tangent function \tan^{-1}x. Here \tan^{-1}(x,y) is the inverse tangent function that takes into account position in the coordinate plane; look up "atan2" for more information)

So we have

z=-1-\sqrt3\,i=2e^{-2\pi/3\,i}

Then

z^6=2^6\left(e^{-2\pi/3\,i}\right)^6=64e^{-4\pi\,i}=64

so that a=64 and b=0.

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Given: measure of angle KJ = 124°, measure of IC =38°<br><br> Find: m∠CQJ, m∠LIJ.
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Part 1) m

Part 2) m

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Part 1) Find the measure of angle CQJ

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m

arc\ IK+arc\ CJ=360\°-(arc\ IC+arc\ KJ)

substitute the values

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m

Part 2) Find the measure of angle LIJ

step 1

Find the measure of angle IJL

we know that

The inscribed angle is half that of the arc it comprises.

m

substitute the values

m

step 2

Find the measure of angle ILJ

we know that

The measurement of the external angle is the semi-difference of the arcs it encompasses.

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step 3

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Remember that the sum of the interior angles of a triangle must be equal to 180 degrees

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