Complex numbers with trig Let z=2e10πi21 and w=eπi7. Then if (z−w)6=reiθ, where r≥0 and 0≤θ&lt;2π what is the ordered pair (r,θ)?

z−w=2e10πi21−eπi7=eπi7(2eπi3−1) But eπi3=12+32i⇒2eπi3−1=3i, so (z−w)6=e6πi7(3i)6=−27eπi7

Consider z−w=eπi7(2eπi3−1) Next 2eπi3−1=2(cos⁡π3+isin⁡π3)−1=i3 Therefore, as π2+π7=9π7 z−w=3e9πi7

z−w=2e10πi21−eπi7=eπi7(2eπi3−1) But eπi3=12+32i⇒2eπi3−1=3i, so (z−w)6=e6πi7(3i)6=−27eπi7

Consider z−w=eπi7(2eπi3−1) Next 2eπi3−1=2(cos⁡π3+isin⁡π3)−1=i3 Therefore, as π2+π7=9π7 z−w=3e9πi7

Expert Answer to "Complex numbers with trig Let z=2e10πi21 and..."

Selena Cowan

2022-01-28

Complex numbers with trig
Let

z = 2 e^{10 π \frac{i}{21}} and w = e^{π \frac{i}{7}}

. Then if

{(z - w)}^{6} = r e^{i θ}

, where

r \geq 0 and 0 \leq θ < 2 π

what is the ordered pair

(r, θ)

Troy Sutton

Beginner2022-01-29Added 13 answers

z - w = 2 e^{10 π \frac{i}{21}} - e^{π \frac{i}{7}} = e^{π \frac{i}{7}} (2 e^{π \frac{i}{3}} - 1)

But

e^{π \frac{i}{3}} = \frac{1}{2} + \frac{\sqrt{3}}{2} i \Rightarrow 2 e^{π \frac{i}{3}} - 1 = \sqrt{3} i

, so

{(z - w)}^{6} = e^{6 π \frac{i}{7}} {(\sqrt{3} i)}^{6} = - 27 e^{π \frac{i}{7}}

mihady54

Beginner2022-01-30Added 13 answers

Consider

z - w = e^{π \frac{i}{7}} (2 e^{π \frac{i}{3}} - 1)

2 e^{π \frac{i}{3}} - 1 = 2 (\cos \frac{π}{3} + i \sin \frac{π}{3}) - 1 = i \sqrt{3}

Therefore, as

\frac{π}{2} + \frac{π}{7} = 9 \frac{π}{7}

z - w = \sqrt{3} e^{9 π \frac{i}{7}}

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