solving for x in complex numbers
asked Jun 1, 2016 in Other Math Topics by Lindiwe

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2 Answers

????? wot du yu meen bi "underline bar", then 243?????
answered Jun 1, 2016 by muneepenee

I read this as 3x^8-243=0; 3x^8-3^5=0; 3x^8=3^5; x^8=3^4.

(x^2)^4=3^4; (x^2)^2=±3^2, so (x^2)^2=3^2 and x^2=3; or (x^2)^2=-3^2 and x^2=3i.

So x=±√3; or x=±√3i.

To find the square root of i let y=a+ib where a and b are real, then y^2=a^2+2abi-b^2=i.

Therefore 2ab=1 and a^2-b^2=0 by equating real and imaginary parts. So a=±b and 2a^2=1 or 2a^2=-1. We can reject the latter because a and b are real, so a=b=1/√2 or √2/2, which is the same thing. Therefore √i=√2(1+i)/2.

So x=±(1+i)√6/2.

The roots are x=-√3, √3, (1+i)√6/2, -(1+i)√6/2.


Let's see if all these roots fit.

x^2=3; x^2=3(1+i)^2/2=3(2i)/2=3i; x^4=9 or -9; x^8=81.

3x^8=243 and 243-243=0, so the roots check out OK.

answered Jun 1, 2016 by Rod Top Rated User (429,320 points)

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