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Airforce X Y / Indian Navy SSR Exam  >  Airforce X Y / Indian Navy SSR Tests  >  Test: Complex Number- 1 - Airforce X Y / Indian Navy SSR MCQ

Test: Complex Number- 1 - Airforce X Y / Indian Navy SSR MCQ


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5 Questions MCQ Test - Test: Complex Number- 1

Test: Complex Number- 1 for Airforce X Y / Indian Navy SSR 2024 is part of Airforce X Y / Indian Navy SSR preparation. The Test: Complex Number- 1 questions and answers have been prepared according to the Airforce X Y / Indian Navy SSR exam syllabus.The Test: Complex Number- 1 MCQs are made for Airforce X Y / Indian Navy SSR 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Complex Number- 1 below.
Solutions of Test: Complex Number- 1 questions in English are available as part of our course for Airforce X Y / Indian Navy SSR & Test: Complex Number- 1 solutions in Hindi for Airforce X Y / Indian Navy SSR course. Download more important topics, notes, lectures and mock test series for Airforce X Y / Indian Navy SSR Exam by signing up for free. Attempt Test: Complex Number- 1 | 5 questions in 10 minutes | Mock test for Airforce X Y / Indian Navy SSR preparation | Free important questions MCQ to study for Airforce X Y / Indian Navy SSR Exam | Download free PDF with solutions
Test: Complex Number- 1 - Question 1
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If a,b (a≠b), are the real roots of the equation (k + 1)(x2 + x + 1)2 + (k - 1)(x4 + x2 + 1) = 0, k ≠ 1, 0. 

Then the product of the roots is

Detailed Solution for Test: Complex Number- 1 - Question 1

Since the equation (x2 +  x + 1) = 0 . oes not have any real roots, the roots of the original equation will be the root of the equation (kx2 + x + k) = 0
 

Hence product of the roots = k/k = 1

Test: Complex Number- 1 - Question 2
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Polar form of a complex number is

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Test: Complex Number- 1 - Question 3
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|z1 + z2 | =

Detailed Solution for Test: Complex Number- 1 - Question 3

|z1 + z2|= .
|z1| + |z2|= .

We have to prove that

 is true.
Square both sides.

Square both sides again.
2x1x2y1y2 ≦ x12y22 + y12x22 and we get
0 ≦ (y1x2 - x1y2)2.
It is true because x1, x2, y1, y2 are all real.

Test: Complex Number- 1 - Question 4
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|z1 - z2 | =
Test: Complex Number- 1 - Question 5
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a2 + b2
Detailed Solution for Test: Complex Number- 1 - Question 5

(a + ib) x (a - ib)

= a2 - i2b2

we know i2=-1

so, equation becomes a2 -(-1)b2 =

a2 + b2

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