Mindblank - how would i find the sq root of 'i'. (1 Viewer)

[amdspotter]

^^^
i tried equating i which can be written as 0+i with x^2+2xyi-y^2 but dont think this works

 

[yanujw]

You were on the right track with equating parts;


Suggesting that the Real and Imaginary parts of your root will be equal or opposite in sign.




Hence,

This can be verified by thinking about roots of a number with modulus 1 geometrically. i = cis(pi/2), so the roots are cis(pi/4) and cis(pi/4 - pi)

 

[amdspotter]

You were on the right track with equating parts;


Suggesting that the Real and Imaginary parts of your root will be equal or opposite in sign.




Hence,

This can be verified by thinking about roots of a number with modulus 1 geometrically. i = cis(pi/2), so the roots are cis(pi/4) and cis(pi/4 -
)

ah ok this makes sense thx. also how would i approach this q as well:
this needs to be proved.

 

[ExtremelyBoredUser]

ah ok this makes sense thx. also how would i approach this q as well: View attachment 34379
this needs to be proved.

Do sum of roots and multiplication of roots to prove that b,c are real. It should become apparent.

 

[CM_Tutor]

The factorised form of the equation must be . By expanding the LHS and equating coefficients, you should find that and must be real so long as is real.

 

[amdspotter]

Ah ok thanks @CM_Tutor @ExtremelyBoredUser

 

[5uckerberg]

^^^
i tried equating i which can be written as 0+i with x^2+2xyi-y^2 but dont think this works

Another way around this is since you have to find the square root of i. A strong note would be since the length of i is 1 and that the equation of the real parts is 0 giving us


This is just simultaneous equations right.
There it becomes

Then find y which is the same
Thus your two roots are