Complex Numbers and Quadratic Equations

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New answer posted•a year ago
New answer posted•a year ago
S
a year ago

Given equation

⇒ c o s x 1 + s i n x = 2 | s i n x | c o s x | c o s 2 x − s i n 2 x |

⇒ | 1 − 2 s i n 2 x | = 2 | s i n x | ( 1 + s i n x )

Put sin x = t

then equation,

| 1 − 2 t 2 | = 2 | t | ( 1 + t ) , t ∈ ( − 1 , 1 ) − { ± 1 2 }

Case I : For ≥ 1 2

the equation has no solution

Case II : For 0 ≤ t < 1 2

Equation t = 5 − 1 4 ⇒ x = 1 8 °

Case III : F o r     − 1 2 < t < 0

Equation t = − 1 2 ⇒ x = − 3 0 °

Case IV : For T < − 1 2

Equation t = 5 + 1 4 ⇒  x = 54°

Sum of solutions = 18° - 30° - 54° = 66°

...Read more

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New answer posted•a year ago
S
a year ago

S = 2 1 + ∑ k = 1 2 1 ( z k + 1 z k ) 3 = ∑ ( z k + z ¯ k ) 3 = 8 ∑ ( R e ( z k ) ) 3

( A s | z | = 1 ⇒ 1 z = z ¯ )

S = 2 ∑ k = 1 2 1 c o s k π + 6 ∑ k = 1 2 1 c o s k π 3 + 2 1

= 2 ( − 1 ) + 6 R e ( α + α 2 + α 3 ) + 2 1

Where α = e i ( 2 π / 6 )

= 2 ( − 1 ) + 6 ( 1 2 − 1 2 − 1 ) + 2 1

= − 8 + 2 1 = 1 3

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New answer posted•a year ago
New answer posted•a year ago
New answer posted•a year ago
S
a year ago

  2 − 4 = − 1 2

y – 4 = 2 (x – 3)

y = 2x – 2

x2 + (2x – 2)2 = 25

  5 x 2 − 8 x − 2 1 = 0

z ( − 7 5 , − 2 4 5 )              

               

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New answer posted•a year ago
S
a year ago

|z−1z|=2

|z|max=?

|z−12|≥||z|−1|z||

2≥|r−1r|

0 ≤r2+2r−1&r2−2r−1≤0

r=−2±82 r=2±82

=−1±2 =1±2

r ≥2−1&0≤r≤1+2

2−1≤r≤2+1

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New answer posted•a year ago
S
a year ago

z ¯ 1 = i z ¯ 2 , a r g ( z 1 z 2 ) = π

⇒ z 1 = − z 2 , a r g ( z 1 z 2 | z 2 | 2 ) = π

⇒ a r g ( z 1 z 2 ) = π

2 0 − π 2 = π

θ = 3 π 4

θ − π 2 = π 4

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New answer posted•a year ago
S
a year ago

D = 0 ⇒ b 2 = 1 5 a , α = 2 b ± 0 2 a = b a  

  b b 2 1 5 = 1 5 b

α + β = 2 b α 2 + β 2 = 4 b 2 − 4 2

α β = 2 1

a ( x − α ) 2 = a ( x 2 − 2 α x + α 2 ) = a x 2 − 2 b x + 1 5

a = 3                    a = -3

b = 7                    a = -7

a2 + b2 = 9 + 49 = 58

 &n

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New answer posted•a year ago
S
a year ago

l i m n ? ? ( n 2 ? n ? 1 + n ? + ? ) = 0

= l i m n ? ? [ 1 ? 1 n ? 1 n 2 + ? + ? n ] = 0

? ? = ? 1

Now, 8 ( ? + ? ) = 8 ( ? 1 2 ) = ? 4

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