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CBSE

Gujarati JEE Mathematics : સંકર સંખ્યાઓ

Multiple Choice Questions

51. જો $bold alpha bold space bold equals bold space bold cos bold space open parentheses fraction numerator bold 8 bold pi over denominator bold 11 end fraction close parentheses bold space bold plus bold space bold i bold space bold sin bold space open parentheses fraction numerator bold 8 bold pi over denominator bold 11 end fraction close parentheses$ તો, $bold Re bold space bold left parenthesis bold space bold alpha bold space bold plus bold space bold alpha to the power of bold 2 bold space bold plus bold space bold alpha to the power of bold 3 bold space bold plus bold space bold alpha to the power of bold 4 bold space bold plus bold space bold alpha to the power of bold 5 bold right parenthesis bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold space bold.$

1
0
$bold minus bold 1 over bold 2$
$bold 1 over bold 2$

52. $bold sum from bold k bold equals bold 1 to bold 6 of bold space open parentheses bold sin open parentheses fraction numerator bold 2 bold kπ over denominator bold 7 end fraction close parentheses bold plus bold space bold i bold space bold cos bold space open parentheses fraction numerator bold 2 bold kπ over denominator bold 7 end fraction close parentheses close parentheses bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold space bold.$

-i
1
0
-1

53. z અને w એ શૂન્યેતર સંકર સંખ્યાઓ છે. જો |zw| = 1 અને argz - argw = $bold pi over bold 2$ તો $bold z with bold bar on top bold space bold w bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold. bold. bold. bold. bold. bold space$

-1
-i
i
1

54. જો |z| = 1 અને z ≠ ± 1 તો કોઈ પણ સંકર સંખ્યા z માટે $fraction numerator bold z over denominator bold 1 bold space bold minus bold space bold z to the power of bold 2 end fraction$ એ = ....... .

Y-અક્ષ પર હોય.
ઉંગમબિંદુમાંથી પસાર ન થતી રેખા પર હોય.
X-અક્ષ પર હોય.
$bold vertical line bold z bold vertical line bold space bold equals bold space square root of bold 2$

55.

જો સંકર સંખ્યા $bold alpha bold space bold left parenthesis bold space bold alpha bold space bold not equal to bold space bold 1 bold right parenthesis$ એ સમીકરણ $bold z to the power of bold p bold plus bold q end exponent bold space bold minus bold space bold z to the power of bold minus bold space bold minus bold space bold z to the power of bold q bold space bold plus bold space bold 1 bold space bold equals bold space bold 0$ જ્યાં p, q ભિન્ન અવિભાજ્ય સંખ્યા છે, નું બીજ હોય તો

1 + α + α² + ... + α^p = 0
1 + α + α² + ... + α^q = 0 અને 1 + α + a² + ... + a^q-1 = 0
1 + α + a² + ... α^p-1 = 0 અને 1 + α + a² + ... + a^q-1 = 0
પરંતુ બંને એક સાથે ન ઉદભવે

56. જો |z| = 1 અને arf (z) = તો arg $open parentheses fraction numerator bold 1 bold plus bold z over denominator bold 1 bold plus bold z end fraction close parentheses bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold space bold.$

θ
$bold pi bold minus bold theta$
-θ
$bold pi over bold 2 bold minus bold theta$

57. જો f(z) = z⁴ - 8z³ + 4z² + 4z + 39 અને f (3 + 2i) = a + ib તો a:b = .........

-1 : 8
1 : 8
1 : 4
-1 : 4

58. જો સંકર સંખ્યા z માટે |z - 3 - 2i| ≤ 2, તો |2z - 6 + 5i| નું મહત્વ મૂલ્ય ....... છે.

59. z અને w એ સંકર સંખ્યાઓ છે. જો $bold z with bold bar on top bold space bold plus bold space bold i bold space bold w with bold bar on top bold space bold equals bold space bold 0$ અને arg (zw) = $bold pi$ તો argz = ............ .

$fraction numerator bold 5 bold pi over denominator bold 4 end fraction$
$fraction numerator bold minus bold pi over denominator bold 4 end fraction$
$bold pi over bold 4$
$bold pi over 2$

fraction numerator bold minus bold pi over denominator bold 4 end fraction

Tips: -

ધારો કે z = re^iθ આથી arg z + arg w = $straight pi$ . આથી $bold arg bold space bold w bold space bold equals bold space bold pi bold space bold minus bold space bold theta$

$bold therefore bold space bold w bold space bold equals bold space bold r subscript bold 1 bold r to the power of bold i bold left parenthesis bold pi bold minus bold theta bold right parenthesis end exponent$ જ્યાં r₁ = |w|

હવે, $bold z with bold bar on top bold space bold plus bold space bold i bold w with bold bar on top bold space bold equals bold space bold 0$

$bold therefore bold space bold re to the power of bold minus bold iθ end exponent bold space bold plus bold space bold ir subscript bold 1 bold e to the power of bold i bold left parenthesis bold theta bold minus bold pi bold right parenthesis end exponent bold space bold equals bold space bold 0 bold r bold space bold left parenthesis bold cos bold space bold theta bold space bold minus bold space bold i bold space bold sin bold space bold theta bold right parenthesis bold space bold plus bold space bold ir subscript bold 1 bold space bold left square bracket bold cos bold space bold left parenthesis bold pi bold minus bold theta bold right parenthesis bold space bold minus bold space bold i bold space bold sin bold space bold left parenthesis bold pi bold minus bold theta bold right parenthesis bold right square bracket bold space bold equals bold space bold 0 bold space bold therefore bold space bold r bold space bold left parenthesis bold cos bold space bold theta bold space bold minus bold space bold i bold space bold sin bold space bold theta bold right parenthesis bold space bold plus bold space bold r subscript bold 1 bold space bold left square bracket bold minus bold i bold space bold cos bold space bold theta bold space bold plus bold space bold sin bold space bold theta bold right parenthesis bold space bold equals bold space bold 0$

વાસ્તવિક ભાગ તથા કાલ્પનિક ભાગ સરખાવતાં,

r cos θ + r₁ sin θ = 0, -r sin θ - r₁ cos θ = 0

$bold therefore bold space bold r over bold r subscript bold 1 bold space bold equals bold space bold minus bold tan bold space bold theta bold space bold equals bold space bold r subscript bold 1 over bold r bold.$ આથી r² = r₁² આથી r = r₁

$bold therefore bold space bold tan bold space bold theta bold space bold equals bold space bold minus bold 1$ આથી $bold argz bold space bold equals bold space fraction numerator bold minus bold pi over denominator bold 4 end fraction$