Question 19 of 28intermediate🔧 ApplyNumerical2 marks

If z1=2−i,z2=1+iz_1 = 2 - i, z_2 = 1 + i, find ∣z1+z2+1z1−z2+1∣\left| \frac{z_1 + z_2 + 1}{z_1 - z_2 + 1} \right|.

Correct Answer

2\sqrt{2}

Exercise: Miscellaneous Exercise on Chapter 4 | Q: 5 | (Page 10)
For More Understanding

Explanation

This question tests the understanding of modulus of complex numbers and the property that the modulus of a quotient equals the quotient of moduli. Students need to first compute the numerator and denominator expressions by adding/subtracting complex numbers, then apply the modulus property.

Solution Steps

  1. Step 1: Calculate the numerator z1+z2+1z_1 + z_2 + 1.

    z1+z2+1=(2−i)+(1+i)+1=2−i+1+i+1=4z_1 + z_2 + 1 = (2 - i) + (1 + i) + 1 = 2 - i + 1 + i + 1 = 4

  2. Step 2: Calculate the denominator z1−z2+1z_1 - z_2 + 1.

    z1−z2+1=(2−i)−(1+i)+1=(2−i)+(−1−i)+1=2−i−1−i+1=2−2iz_1 - z_2 + 1 = (2 - i) - (1 + i) + 1 = (2 - i) + (-1 - i) + 1 = 2 - i - 1 - i + 1 = 2 - 2i

  3. Step 3: Apply the modulus property ∣z1z2∣=∣z1∣∣z2∣\left| \frac{z_1}{z_2} \right| = \frac{|z_1|}{|z_2|}.

    ∣42−2i∣=∣4∣∣2−2i∣\left| \frac{4}{2 - 2i} \right| = \frac{|4|}{|2 - 2i|}

  4. Step 4: Calculate individual moduli.

    ∣4∣=4|4| = 4 ∣2−2i∣=22+(−2)2=4+4=8=22|2 - 2i| = \sqrt{2^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}

  5. Step 5: Compute the final answer.

    ∣4∣∣2−2i∣=422=22=2\frac{|4|}{|2 - 2i|} = \frac{4}{2\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}
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