Question

Let w=2+i and z be nonzero complex number. If the argument of z is pi/4 and |x-w| =sqrt(5) , what is | z |^2

Answer 1

Explanation:

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|z|^2 = |z|*|z| = (|z|)^2

Since the argument of z is pi/4, z can be written as z=r*(cos(pi/4)+i*sin(pi/4)) where r is the magnitude of z.

We also know that |x-w| =sqrt(5).

This means that |z-w| = sqrt(5).

Substituting the expression for z, we get:

|r*(cos(pi/4)+i*sin(pi/4))-w| = sqrt(5)

Simplifying this equation, we get:

|r-2-i| = sqrt(5)

Squaring both sides, we get:

(r-2-i)(r-2+i) = 5

Simplifying, we get:

r^2-4+i(2r-2)=5

Equating the real and the imaginary parts, we get:

r^2-4 = 5

and

2r-2=0

Solving these equations simultaneously, we get:

r = sqrt(9)

Therefore,

|z| = sqrt(9)

Hence,

|z|^2 = (sqrt(9))^2 = 9