Question

pls answer the ques of complex no

Answer 1

Answer:

$\bold\red{(b)\:the\: imaginary\:axis}$

Step-by-step explanation:

Given,

$| {z}^{2} - 1| = | {z}^{2} | + 1$

Let,

z = x + iy

Then,

we get,

$= > | {(x + iy) }^{2} - 1 | = { |x + iy| }^{2} + 1 \\ \\ = > | {x}^{2} + {i}^{2} {y}^{2} + i(2xy) | = {x}^{2} + {y}^{2} + 1 \\ \\ = > |( {x}^{2} - {y}^{2} - 1) + i(2xy) | = {x}^{2} + {y}^{2} + 1 \\ \\ = > {( {x}^{2} - {y}^{2} - 1) }^{2} + {(2xy)}^{2} = {( {x}^{2} + {y}^{2} + 1) }^{2} \\ \\ = > {x}^{4} + {y}^{4} + 1 + 2 {x}^{2} {y}^{2} + 2 {y}^{2} - 2 {x}^{2} \\ = {x}^{4} + {y}^{4} + 1 + 2 {x}^{2} {y}^{2} + 2 {y}^{2} + 2 {x}^{2} \\ \\ \\ = > 4 {x}^{2} = 0 \\ \\ = > x = 0$

Therefore,

z = iy

Hence,

z lies on (b)the imaginary axis.

Answer 2

Answer:

❤❤Appy Diwali Virus❤

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*sorry for Answering....

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