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  • The locus of the point whose position vector is given by <img alt="\{a+i b)^3+(b+i a\}^3" src="https://entrancecorner.oncodecogs.com/gif.latex?%5C%7Ba&plus;i%20b%29%5E3&plus;

The locus of the point whose position vector is given by \{a+i b)^3+(b+i a\}^3 (where a, b are real parameters) is
 

 

Option: 1

y=x
 


Option: 2

y=-x
 


Option: 3

y=m x, n \in R
 


Option: 4

not defined


Answers (1)

best_answer

\begin{aligned} & \text{Let } a=r \cos \theta b=r \sin \theta \\ & \text{ Thus } (a+i b)^3+(b+ia)^3 \\ & \left.=r^5\{\cos \theta+i \sin \theta)^5+(\sin \theta+i \cos \theta)^5\right\} \\ & =r^5\left[(\cos 5 \theta+i \sin 5 \theta]+\left\{\cos \left(\frac{\pi}{2}-\theta\right)+i \sin \left(\frac{\pi}{2}-\theta\right)\right\}^5\right] \\ & =r^5\left[[\cos 5 \theta+i \sin 5 \theta)+\cos 5\left(\frac{\pi}{2}-\theta\right)+i \sin 5\left(\frac{\pi}{2}-\theta\right)\right] \\ & =r^5[(\cos 5 \theta+i \sin 5 \theta)(1+i)] \end{aligned}
This is a complex number whose real and imeginary parts are equal. So, locus of such a point will be y=x.

Posted by

Rakesh

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