Question

a function is defined on the complex numbers by where and are positive numbers. this function has the property that for each complex number , is equidistant from both and the origin. given that , find

Answer 1

The value of p+q = 403,For the given complex number a+bi and $b^{2} =\frac{p}{q}$

where p and q are co-primes

F(z)= (a+ib)z⇒this is equidistant from "0" and "z"

Given  modulus of complex number (a+ib) = 10 ; $b^{2} =\frac{p}{q}$ p and q ∈Z

G.C.D of ( p and q)=1

(a+ib)z equidistant from "0" and "z"

$&\Rightarrow|(a+i b) z-z|=|(a+i b) \bar{z}-0|\\&|z(a+i b-1)|=|(a+i b) \bar{z}|\\&[\bar{z}||(a-1)+i b|=| z|(a+i b)|\\&|a-1+i b|=|a+i b|\\&\sqrt{(a-1)^2+b^2}=\sqrt{a^2+b^2}\\&|a+i b|=10 \quad a^2+1-2 a+b^2=x^2+b^2\\&\sqrt{a^2+b^2}=10\\&a=1 / 2\\&a^2+b^2=100\\&b^2=100-\frac{1}{4}$

$b^{2} =\frac{399}{4}$

p = 399 and q= 4

p+q= 399+4

p+q=403

Hence the value of p+q = 403

Complete question:A function f is defined on the complex number by f (z) = (a + bi)z, where 'a' and 'b' are positive numbers. This function has the property that the image of each point in the complex plane is equidistant from that point and the origin. Given that |a+bi|=8 and that$b^{2} =\frac{p}{q}$  where p and q are coprime. Find the value of (p+q)

Learn more about complex numbers here:

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