if |z-5i|/|z+5i| = 1, then show that z is a real number

Let z=a+ibz-5i=a+ib-5z+5i=a+ib+5z-5iz+5i=1a2+b-52a2+b+52=1a2+b-52=a2+b+52Squaring on both sides , we geta2+b-52=a2+b+52b2-10b+25=b2+10b+25-20b=0        b=0So, z=a , which is a real number.

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