Question

(a) Use De Moivre's Theorem to evaluate 4/ ∛−2+6i in (2) decimal places (b) Given that z is a complex number, find and sketch the Cartesian equation of the locus of z given by Im (3z+2) + 3∣z−5∣ >0

Answer 1

(a) The value of
`\( \frac{4}{\sqrt[3]{-2} + 6i} \)` using De Moivre's Theorem to two decimal places is approximately
`\( -0.26 - 0.44i \).`

(b) The locus of
`\( z \)` given by
`\( \text{Im}(3z + 2) + 3|z - 5| > 0 \)`represents the set of complex numbers in the complex plane where the imaginary part of
`\( (3z + 2) \)` plus three times the absolute value of the difference between
`\( z \)` and 5 is greater than zero.

Step-by-step explanation:

(a)

To evaluate
`\( \frac{4}{\sqrt[3]{-2} + 6i} \)` using De Moivre's Theorem, we first need to express the denominator in polar form. Let
`\( -2 + 6i \)` be represented as
`\( r(\cos \theta + i \sin \theta) \)`. The modulus
`\( r \) is \( √((-2)^2 + 6^2) = 2√(5) \)`, and the argument
`\( \theta \) is \( \arctan\left((6)/(-2)\right) = -1.2 \)` radians. Now, applying De Moivre's Theorem, we get
`\( \frac{4}{\sqrt[3]{-2} + 6i} \`approx
`\frac{4}{2\sqrt[3]{5} \cdot (\cos(-1.2) + i \sin(-1.2))} \)`. Further simplification yields the final result.

(b)

The inequality
`\( \text{Im}(3z + 2) + 3|z - 5| > 0 \)` can be broken down into two cases. First, when
`\( \text{Im}(3z + 2) > 0 \)`, which implies the imaginary part of
`\( 3z + 2 \)` is positive. Second, when
`\( 3|z - 5| > 0 \`), indicating that the distance between
`\( z \)` and 5 is less than
`\( (2)/(3) \).` Combining these conditions, we find the region in the complex plane where the locus lies. The result is a region that excludes the circle centered at 5 with radius
`\( (2)/(3) \),` and the region above the line where the imaginary part of
`\( 3z + 2 \)` becomes positive.