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Exercise 2.5 · Q8

Q.If the area of the triangle formed by the vertices z,izz, iz, and z+izz+iz is 5050 square units, find the value of ∣z∣|z|.

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Multiplying zz by ii rotates it 90°90° about the origin, so the triangle with vertices z,iz,z+izz,iz,z+iz is right-angled with two equal legs of length ∣z∣|z|; equate 12∣z∣2\tfrac12|z|^2 to the given area.

Step 1. Label vertices and compute the three side lengths. Let A=z, B=iz, C=z+izA=z,\ B=iz,\ C=z+iz.

BC→=C−B=(z+iz)−iz=z⇒∣BC∣=∣z∣\overrightarrow{BC}=C-B=(z+iz)-iz=z \Rightarrow |BC|=|z|.

CA→=A−C=z−(z+iz)=−iz⇒∣CA∣=∣−iz∣=∣i∣∣z∣=∣z∣\overrightarrow{CA}=A-C=z-(z+iz)=-iz \Rightarrow |CA|=|-iz|=|i||z|=|z|.

AB→=B−A=iz−z=z(i−1)⇒∣AB∣=∣z∣ ∣i−1∣=∣z∣2\overrightarrow{AB}=B-A=iz-z=z(i-1) \Rightarrow |AB|=|z|\,|i-1|=|z|\sqrt2. …

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