Question

Please help answer this question

Answer 1

Answer:

Step-by-step explanation:

The locus is a circle centered at (2,0) with a radius of more than 3, meaning the equation is $\boxed{(x-2)^{2}+y^{2} > 9}$

The graph is shown in the attached image.

Answer 2

a. The locus |z - 2| > 3 is [(x - 2)² + y²] > 9

The locus |z - 2| > 3 is the region outside the circle [(x - 2)² + y²] = 9

b. Find the argand diagram in the attachment

a. How to find the locus?

Since z = x + iy, and we need to find the locus of |z - 2| > 3.

So, z - 2 = x + iy - 2

= (x - 2) + iy

The modulus of a complex number

For a complex number z = x + iy, its modulus |z| = √(x² + y²)

So, the modulus of |z - 2| = √[(x - 2)² + y²]

The locus of |z - 2| > 3

So, the locus |z - 2| > 3

√[(x - 2)² + y²] > 3.

Squaring both sides, we have

(√[(x - 2)² + y²])² > 3².

[(x - 2)² + y²] > 9

Comparing this to the equation of a circle with center (h,k)

(x - h)² + (y - k)²] = r²

So, r² = 9 ⇒ r = √9 = 3 and (h, k) = (2,0)

The locus |z - 2| > 3 is [(x - 2)² + y²] > 9

So, the locus |z - 2| > 3 is the region outside the circle [(x - 2)² + y²] = 9

b. Find the argand diagram in the attachment

Learn more about locus here:

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