Question

# Let z be a complex number satisfying equation zp=¯zq, where p,q ϵ N, then

A
If p=q, then number of solutions of equation will be infinite
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B
If p q, then number of solutions of equation will be p+q
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C
If p q, then number of solutions of equation will be p+q+1
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D
If p=q, then number of solutions of equation will be finite
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Solution

## The correct option is C If p≠ q, then number of solutions of equation will be p+q+1Given: zp=¯zq⇒ zp=¯zq (∵ qϵN) If p=q, then equation zp=¯zq has infinite number of solutions because any zϵR will satisfy it. If p≠ q, let p>q, then zp=¯zq⇒|z|p=|z|q ⇒|z|q(|z|p−q−1)=0 ⇒|z|=0 or |z|=1 |z|=0⇒ z=0+i0 |z|=1⇒ z=eiθ ∵ zp=¯zq ⇒ epθ i=e−qθ i ⇒ e(p+q)θ i=1 ⇒ z=11p+q ∴ The number of solutions is p+q+1. Hence, option (A) and (C) are correct.

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