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qaws [65]
3 years ago
13

Describe the set of points z in the complex plane that satisfy the given equation; |z-2|=Re(z).

Mathematics
1 answer:
OLEGan [10]3 years ago
3 0

Answer:

Parabola with vertex at point (1,0) that goes to the right (see attached diagram).

Step-by-step explanation:

Let the complex number z be

z=x+iy,

then

z-2=x+iy-2=(x-2)+iy\\ \\Re\ (z-2)=x-2\\ \\Im\ (z-2)=y\\ \\|z-2|=\sqrt{Re^2\ (z-2)+Im^2\ (z-2)}=\sqrt{(x-2)^2+y^2}

and

Re\ z=x

Thus,

\sqrt{(x-2)^2+y^2}=x

Square it:

(x-2)^2+y^2=x^2\\ \\x^2-4x+4+y^2=x^2\\ \\-4x+4+y^2=0\\ \\y^2=4x-4\\ \\y^2=4(x-1)

This is the equation of parabola with vertex at point (1,0) that goes to the right.

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\sf\Huge\boxed{C.40}

Step-by-step explanation:

Hello There!

Remember: sum of interior angles of a triangle = 180

so to find x we use this equation

180 = 90 + 7x + 5 + 9x + 5 ( the little square in the triangle indicates that the angle is a right angle. right angles have a measure of 90 so that's where the 90 came from.)

now we solve for x

step 1 combine like terms

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2 years ago
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