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Exercise 2.9 · Q3

Q.The area of the triangle formed by the complex numbers z,izz, iz, and z+izz+iz in the Argand's diagram is

(1) 12∣z∣2\dfrac12|z|^2
(2) ∣z∣2|z|^2
(3) 32∣z∣2\dfrac32|z|^2
(4) 2∣z∣22|z|^2
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✓ Free question

Multiplying zz by ii rotates it 90∘90^\circ about the origin, so two of the triangle's sides are perpendicular and equal in length ∣z∣|z| — giving a right isosceles triangle whose area is half the product of its legs.

Step 1. Identify the three vertices as vectors from the origin. A=zA=z, B=izB=iz, C=z+izC=z+iz.

Step 2. Compute the side AC→=C−A\overrightarrow{AC}=C-A.

AC→=(z+iz)−z=iz⇒∣AC→∣=∣iz∣=∣i∣∣z∣=∣z∣.\overrightarrow{AC}=(z+iz)-z=iz \quad\Rightarrow\quad |\overrightarrow{AC}|=|iz|=|i||z|=|z|.

Step 3. Compute the side CB→=B−C\overrightarrow{CB}=B-C.

CB→=iz−(z+iz)=−z⇒∣CB→∣=∣−z∣=∣z∣.\overrightarrow{CB}=iz-(z+iz)=-z \quad\Rightarrow\quad |\overrightarrow{CB}|=|-z|=|z|.

Step 4. Show the angle at CC (between CA→=−iz\overrightarrow{CA}=-iz and CB→=−z\overrightarrow{CB}=-z) is 90∘90^\circ. Since CA→=−iz\overrightarrow{CA}=-iz is exactly CB→=−z\overrightarrow{CB}=-z rotated by 90∘90^\circ (multiplied by ii), the two legs CACA and CBCB are perpendicular. So triangle ABCABC is right-angled at CC with legs CA=∣z∣CA=|z| and CB=∣z∣CB=|z|.

Step 5. Compute the area of the right triangle with two equal legs of length ∣z∣|z|.

Area=12×∣z∣×∣z∣=12∣z∣2.\text{Area}=\frac12\times|z|\times|z|=\frac12|z|^2.

✓Final answer

Option (1): 12∣z∣2\dfrac12|z|^2.

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