Question

যদি
$x = \frac{1}{ \sqrt{2} } (1 - i)$
তেন্তে দেখুওৱা যে x⁶+x⁴+x²+1=0​

Answer 1

Answer:

Given : x= (1+i) /√2

x² = (1+i)(1+i)/(√2)² = 2i/2 = i

x⁴ = (1+i)(1+i)(1+i)(1+i)/(√2)⁴ = i.i = i² = -1

x⁶ = (1+i)(1+i)(1+i)(1+i)(1+i)(1+i)/(√2)⁶ = (-1).i = -i

So, x⁶ + x⁴ +x² +1 = -i-1+i+1 = 0

Answer 2

Step-by-step explanation:

দিয়া আছে,

$x = \frac{1}{ \sqrt{2} } (1 - i)$

=>√2x=(1-i)

=>(√2x)²=(1-i)²

=>2x²=1-2i+i²

=>2x²= -2i

=>x²= -i

L.H.S.= x⁶+ⅹ⁴+ⅹ²+1

=(x²)³+(x²)²+x²+1

=(-i)³+(-i)²+(-i)+1

=i-1-i+1

=0

proved.......