Given z is a complex number.Let z = x + yi⇒ |z| = |x + yi| and |z|2 = |x + yi|2⇒ |z|2 = x2 + y2 … (1)Now z2 = x2 + y2i2 + 2xyi⇒ z2 = x2 – y2 + 2xyiSo, |z|2 = x2 + y2 = |z|2∴ |z|2 = |z2|

Q. 463.8( 8 Votes )

If z is a complex A. |z²| > |z|²

B. |z²| = |z|²

C. |z²| < |z|²

D. |z²| ≥ |z|²

Class 11th Mathematics - Exemplar 5. Complex Numbers and Quadratic Equations

Answer :

Given z is a complex number.

Let z = x + yi

⇒ |z| = |x + yi| and |z|² = |x + yi|²

⇒ |z|² = x² + y² … (1)

Now z² = x² + y²i² + 2xyi

⇒ z² = x² – y² + 2xyi

So, |z|² = x² + y² = |z|²

∴ |z|² = |z²|

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If z is a complex A. |z2| > |z|2B. |z2| = |z|2C. |z2| < |z|2D. |z2| ≥ |z|2

If z is a complex A. |z²| > |z|²

B. |z²| = |z|²

C. |z²| < |z|²

D. |z²| ≥ |z|²