Complex Numbers 4 Question 11
11.
If $a$ and $b$ are real numbers between 0 and 1 such that the points $z _1=a+i, z _2=1+b i$ and $z _3=0$ form an equilateral triangle, then $a=\ldots$ and $b=$…
$(1990,2 M)$
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Answer:
Correct Answer: 11. $a=b=2 \pm \sqrt{3}$
Solution:
- Since, $z _1, z _2$ and $z _3$ form an equilateral triangle.
$ \begin{aligned} & \Rightarrow \quad z _1^{2}+z _2^{2}+z _3^{2}=z _1 z _2+z _2 z _3+z _3 z _1 \\ & \Rightarrow \quad(a+i)^{2}+(1+i b)^{2}+(0)^{2}=(a+i)(1+i b)+0+0 \\ & \Rightarrow \quad a^{2}-1+2 i a+1-b^{2}+2 i b=a+i(a b+1)-b \\ & \Rightarrow \quad\left(a^{2}-b^{2}\right)+2 i(a+b)=(a-b)+i(a b+1) \\ & \Rightarrow \quad a^{2}-b^{2}=a-b \\ & \text { and } \quad 2(a+b)=a b+1 \\ & \Rightarrow \quad(a=b \text { or } a+b=1) \\ & \text { and } \quad 2(a+b)=a b+1 \\ & \text { If } a=b, \quad 2(2 a)=a^{2}+1 \\ & \Rightarrow \quad a^{2}-4 a+1=0 \\ & \Rightarrow \quad a=\frac{4 \pm \sqrt{16-4}}{2}=2 \pm \sqrt{3} \\ & \text { If } \quad a+b=1,2=a(1-a)+1 \Rightarrow a^{2}-a+1=0 \\ & \Rightarrow a=\frac{1 \pm \sqrt{1-4}}{2}, \text { but } a \text { and } b \in R \\ & \therefore \text {Only solution when}\quad a=b \\ & a=b=2 \pm \sqrt{3} \\ & a=b=2-\sqrt{3} \quad[\because a, b \in(0,1)] \end{aligned} $
