Let z1 and z2 be complex numbers such that z1 ≠ z2 and | z1| = | z2 | . If Re (z1) 0 and Im (z2)

Question

Let z1 and z2 be complex numbers such that z1 ≠ z2 and | z1| = | z2 | . If Re (z1) > 0 and Im (z2) < 0 ,then (z1+ z2/z1 - z2) is (A) One (B) real and positive (C) real and negative (D) purely imaginary

Tardigrade Admin · Accepted Answer

Let z1=x1+i y1 and z2=x2+i y2 Re (z1)>0 ⇒ x1>0 and Im (z2) < 0 ⇒ y2 < 0 Given, |z1|=|z2| ⇒ |z1|2=|z22| ⇒ z1 barz1 = z2 barz2 Now, ((z1+z2/z1-z2))+(( overlinez1+z2/z1-z2)) =((z1+z2/z1-z2))+(( barz1+ barz2/ barz1- barz2)) =(z1 barz1+z2 barz1-z1 barz2-z2 barz2+z1 barz1+z1 barz2-z2 barz1+z2 barz2/(z1-z2)( barz1- barz2)) =(2(|z1|2-|z2|2)/(z1-z2)( barz1- barz2))=0 (∵|z1|2=|z2|2) =(z1+z2/z1-z2) is purely imaginary.

Q. Let z1​ and z2​ be complex numbers such that z1​=z2​ and ∣z1​∣=∣z2​∣ . If Re (z1​)>0 and Im(z2​)<0 ,then z1​−z2​z1​+z2​​ is

One

real and positive

real and negative

purely imaginary

Q. Let $z_{1}$ and $z_{2}$ be complex numbers such that $z_{1} \neq = z_{2}$ and $∣ z_{1} ∣ = ∣ z_{2} ∣$ . If Re $(z_{1}) > 0$ and $I m (z_{2}) < 0$ ,then $\frac{z _{1} + z _{2}}{z _{1} - z _{2}}$ is