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

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:

To avoid this verification in future, please log in or register.

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 (486,900 points)
Welcome to, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
79,816 questions
83,634 answers
66,552 users