Question

the modulus of the complex number (1-iroot3/1+i root 3)​

Answer 1

Answer:-

| z | = 1

Step - by - step explanation:-

To find-

Modulus of given complex number .

Solution:-

Firstly solve it,

$let \: \: \: z = \frac{1 - i \sqrt{3} }{1 + i \sqrt{3} } \\ \\ rationalising \: by \:( 1 - i \sqrt{3} ) \\ \\ z = \frac{1 - i \sqrt{3} }{1 +i \sqrt{3} } \times \frac{1 - i \sqrt{3} }{1 - i \sqrt{3} } \\ \\ z = \frac{ {(1 - i \sqrt{3)} }^{2} }{1 - 3 {i}^{2} } \\ \because \: {i}^{2} = - 1 \: \\ \therefore \: \\ \\ z = \frac{ {(1 - i \sqrt{3}) }^{2} }{1 + 3} \\ \\ z = \frac{1 - 3 - i2 \sqrt{3} }{4} \\ \\ z = \frac{ - 2}{4} - \frac{i2 \sqrt{3} }{4} \\ \\ z = \frac{ - 1}{2} - \frac{i \sqrt{3} }{2}$

Compare this to the general form of a complex number ,

$\implies \: z = x + iy \\ \\ after \: \: compairing \: \: we \: get \\ \\ x = \frac{ - 1}{2} \: \: and \: y = \frac{ - \sqrt{3} }{2} \\ \\ \\| z| = \sqrt{ {x}^{2} + {y}^{2} } \\ \\ |z| = \sqrt{ \frac{1}{4} + \frac{3}{4} } \\ \\| z |= \sqrt{ \frac{4}{4} } = 1$

Answer 2

Answer:

Step-by-step explanation:

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