Question

modulus of 1-i whole square​

Answer 1

Given ,

Complex number (z) = (1 - i)²

On simplifying it , we get

➡(1)² + (i)² - 2 × 1 × i

➡1 + (-1) - 2i

➡1 - 1 - 2i

➡-2i

Therefore , the complex number in standard form i.e a + ib is 0 - 2i

Now , the modulus of complex number is given by

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

Thus ,

r = √{(0)² + (-2)²}

r = √{4}

r = 2 { since r > 1 }

Therefore , the modulus of complex number is 2

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Answer 2

Given ,

Complex number (z) = (1 - i)²

On simplifying it , we get

➡(1)² + (i)² - 2 × 1 × i

➡1 + (-1) - 2i

➡1 - 1 - 2i

➡-2i

Therefore , the complex number in standard form i.e a + ib is 0 - 2i

Now , the modulus of complex number is given by

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

Thus ,

r = √{(0)² + (-2)²}

r = √{4}

r = 2 { since r > 1 }

Therefore , the modulus of complex number is 2

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