Question

Let Z_i = r_i (cos q_i + i sinq_i), i = 1, 2, 3 and $\frac{1}{Z_{1}}$+ $\frac{1}{Z_{2}}$+ $\frac{1}{Z_{3}}$= 0. Consider the DABC formed by $\frac{\cos 2\theta_{2} + i\sin 2\theta_{2}}{Z_{2}}$, $\frac{\cos 2\theta_{3} + i\sin 2\theta_{3}}{Z_{3}}$.

Then the complex number 0 lies

• A. On the side BC
• B. Outside the triangle
• C. Inside the triangle
• D. On the side CA

Answer 1

Answer: (C)

Inside the triangle

Answer 2

Sol. $\frac{1}{Z_{1}}$+ $\frac{1}{Z_{2}}$+ $\frac{1}{Z_{3}}$= 0

Ž $\sum_{}^{}\frac{1}{r_{1}(\cos\theta_{1} + ⥂ i\sin\theta_{1})}$= 0

Ž $\sum_{}^{}\frac{\cos\theta_{1} - i\sin\theta_{1}}{r_{1}}$= 0

Ž $\sum_{}^{}\frac{\cos\theta_{1} + i\sin\theta_{1}}{r_{1}}$= 0

Ž $\sum_{}^{}\frac{(\cos\theta_{1} + i\sin\theta_{1})^{2}}{r_{1}(\cos\theta_{1} + i\sin\theta_{1})}$= 0

Ž $\sum_{}^{}\frac{\cos 2\theta_{1} + i\sin 2\theta_{1}}{Z_{1}}$= 0

Ž $\frac { 1 } { 3 }$ $\frac{(\cos 2\theta_{1} + i\sin 2\theta_{1})}{Z_{1}}$= 0

Ž the centroid of the D ABC is the origin

Ž origin lies inside the D.