Complex numbers problem!! (1 Viewer)

[Premus]

Hey

can u guys help me with this question?

If 'x' is real and (x + i) ^4 is imaginary, find the possible values of x, in surd form.

also, do u know if there are solutions to the 2003 CSSA paper?

Thanks!!

 

[edd91]

if x is real, it meens Im(x)=0
that meens x + i = Re(x) + i
Solve (x+i)^4 for x, and then subtract i from the answers?

 

[Premus]

sorry i didnt really understand you....

i get the part where u said: if x is real, it means Im(x)=0
but i dont understand how u get x + i = Re(x) + i
or how you solve for x??

Thanks

 

[edd91]

sorry, looks like i oversimplified the question
i have a feeling of the method but I'd like to see a more confident answer from someone

 

[CM_Tutor]

(x + i)⁴ = x⁴ + 4x³i + 6x²i² + 4xi³ + i⁴ = x⁴ + 4ix³ - 6x² - 4ix + 1

Now, (x + i)⁴ is purely imaginary, then its real part is zero. So, x⁴ - 6x² + 1 = 0
x² = [6 +/- sqrt(36 - 4)] / 2 = 3 +/- 2 * sqrt(2), using the quadratic formula

So, there are four possible values for x, +/- sqrt(3 +/- 2 * sqrt(2))

 

[Premus]

Thanks!!