Test: Complex Number- 1


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5 Questions MCQ Test Mathematics for Airmen Group X | Test: Complex Number- 1

Test: Complex Number- 1 for Airforce X Y / Indian Navy SSR 2023 is part of Mathematics for Airmen Group X 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 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Complex Number- 1 below.
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Test: Complex Number- 1 - Question 1

If a,b (a≠b), are the real roots of the equation (k + 1)(x+ x + 1)+ (k - 1)(x+ x+ 1) = 0, k ≠ 1, 0. 

Then the product of the roots is

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

Since the equation (x+  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

Polar form of a complex number is

Test: Complex Number- 1 - Question 3

|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 ≦ x12y2y12x22 and we get
0 ≦ (y1x2 - x1y2)2.
It is true because x1, x2y1y2 are all real.

Test: Complex Number- 1 - Question 4

|z1 - z2 | =

Test: Complex Number- 1 - Question 5

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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