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Exercise 1.1 · Q48

Q.If a=−12+32ia = -\dfrac12+\dfrac{\sqrt3}{2}i, b=−12−32ib = -\dfrac12-\dfrac{\sqrt3}{2}i then show that a2=ba^2=b and b2=ab^2=a.

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a=−12+32ia=-\dfrac12+\dfrac{\sqrt3}{2}i. Squaring: a2=(−12)2+2(−12)(32i)+(32i)2=14−32i+34i2=14−32i−34=−12−32i=ba^2=\left(-\dfrac12\right)^2+2\left(-\dfrac12\right)\left(\dfrac{\sqrt3}{2}i\right)+\left(\dfrac{\sqrt3}{2}i\right)^2=\dfrac14-\dfrac{\sqrt3}{2}i+\dfrac34 i^2=\dfrac14-\dfrac{\sqrt3}{2}i-\dfrac34=-\dfrac12-\dfrac{\sqrt3}{2}i=b. So a2=ba^2=b. By exactly the sam …

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