Question

Modulus of (1-2i)/(1+2i) is-

Answer 1

Answer:

Given : (1-2i)/(1+2i)

⇒ (1-2i)/(1+2i) × (1-2i)/(1-2i)

⇒ (1-2i-2i+4i²)/(1-2i+2i-4i²)

⇒ (1-4i-4)/(1+4)

⇒ (-3-4i)/(5)

⇒ -3/5 - 4/5 i

Modulus : √(x² + y²)

⇒ √(-3/5)² + (-4/5)²

⇒ √9/25+16/25

⇒ √(9+16)/25

⇒ √25/25

⇒ √1

⇒ 1

Hence, Modulus of (1-2i)/(1+2i) is : 1.

IMPORTANT :

• Negative numbers do not have square roots.
• i = √-1
• i² = -1
• z = a + ib
• a and b are real.
• a is the real part.
• ib is the imaginary part.

Answer 2

Given ,

The complex number is (1-2i)/(1+2i)

Multiplying numerator and denominator by 1 - 2i , we get

$\tt \implies \frac{(1 - 2i)(1 - 2i)}{(1 + 2i)(1 - 2i)}$

$\tt \implies \frac{ {(1 - 2i)}^{2} }{ {(1)}^{2} - {(2i)}^{2} }$

$\tt \implies \frac{ {(1)}^{2} + {(2i)}^{2} - 2(1)(2i) }{1 - 4 (- 1)}$

$\tt \implies \frac{1 + 4( - 1) - 4i}{5}$

$\tt \implies \frac{ - 3 - 4i}{5}$

$\tt \implies - \frac{ 3}{5} - \frac{4i}{5}$

Therefore , the complex number in standard form is -3/5 - 4i/5

On comparing with a + ib , we get

a = -3/5

b = -4/5

Now , the distance between the origin and (a , b) is called modulus of complex number ( z = a + ib)

Mathematically ,

$\boxed{ \tt{ |z| = r = \sqrt{ {(a)}^{2} + {(b)}^{2} }} }$

Thus ,

$\tt \implies |z| = \sqrt{ {( - \frac{3}{5} )}^{2} + {( - \frac{4}{5} )}^{2} }$

$\tt \implies |z| = \sqrt{ \frac{9}{25} + \frac{16}{25} }$

$\tt \implies |z| = \sqrt{ \frac{9 + 16}{25} }$

$\tt \implies |z| = \sqrt{ \frac{25}{25} }$

$\tt \implies|z| = \sqrt{1}$

$\tt \implies |z| = 1$

Therefore , the modulus of given complex number 1

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