The numbers 4, 5, 6, and 7 are on a spinner. You spin the spinner twice. Which calculation proves that landing on an even number
2 answers:
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Given <em>z</em> = 3 + <em>i</em>, right away we can find
(a) square
<em>z</em> ² = (3 + <em>i </em>)² = 3² + 6<em>i</em> + <em>i</em> ² = 9 + 6<em>i</em> - 1 = 8 + 6<em>i</em>
(b) modulus
|<em>z</em>| = √(3² + 1²) = √(9 + 1) = √10
(d) polar form
First find the argument:
arg(<em>z</em>) = arctan(1/3)
Then
<em>z</em> = |<em>z</em>| exp(<em>i</em> arg(<em>z</em>))
<em>z</em> = √10 exp(<em>i</em> arctan(1/3))
or
<em>z</em> = √10 (cos(arctan(1/3)) + <em>i</em> sin(arctan(1/3))
(c) square root
Any complex number has 2 square roots. Using the polar form from part (d), we have
√<em>z</em> = √(√10) exp(<em>i</em> arctan(1/3) / 2)
and
√<em>z</em> = √(√10) exp(<em>i</em> (arctan(1/3) + 2<em>π</em>) / 2)
Then in standard rectangular form, we have
and
We can simplify this further. We know that <em>z</em> lies in the first quadrant, so
0 < arg(<em>z</em>) = arctan(1/3) < <em>π</em>/2
which means
0 < 1/2 arctan(1/3) < <em>π</em>/4
Then both cos(1/2 arctan(1/3)) and sin(1/2 arctan(1/3)) are positive. Using the half-angle identity, we then have
and since cos(<em>x</em> + <em>π</em>) = -cos(<em>x</em>) and sin(<em>x</em> + <em>π</em>) = -sin(<em>x</em>),
Now, arctan(1/3) is an angle <em>y</em> such that tan(<em>y</em>) = 1/3. In a right triangle satisfying this relation, we would see that cos(<em>y</em>) = 3/√10 and sin(<em>y</em>) = 1/√10. Then
So the two square roots of <em>z</em> are
and
Express in simplest a+bi form: Square root of 1b + Square root of -1b
Lets simplify
We know
We cannot simplify
So a + ib form is