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10 months ago

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abhishek gaurav

Contributor-Level 9

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Payal Gupta

Contributor-Level 10

28. To proof, Re (z1z2) = Re z1 Re z2 – Imz1 Imz2

Let z­­1 = x1 + iy1 and z2 = x2 + iy2 be two complex number.

Then, z1.z2 = (x1 + iy1) (x2 + iy2)

=x1x2 + ix1y2 + ix2y1 + i2y1y2

= x1x2 + ix1y2 + ix2y1y1y2            [since, i2 = -1]

= (x1x2y1y2) + i (x1y2 + x2y1)

As, Re (z1z2) = (x1) (x2) – (y1) (y2)

Now, RHS = Re z1 Re z2Imz1Imz2 = x1x2y1y2

Therefore, Re (z1z2) = Rez1Rez2Imz1Imz2

Hence proved.

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abhishek gaurav

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abhishek gaurav

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Payal Gupta

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28.  [i18+ (1i)23]3

[i4?
4+2+1i4?
6+1]3

[? 1+1i]3 [as i4 *k + 2 = –1 and i4 *k + 1 = i]

[? 1+ii2]3

= [–1 – i]3                        [as i2 = –1]

= (-1)3 (1 + i)3

= –1 [13 + i3 + 3 * 1 *i (1 + i)]                        [since, (a + b)3 = a3 + b3 + 3ab (a + b)]

= –1 [1 – i3 + 3i (1 + i)]

= –1 [1 – i3 + 3i + 3i2]

= –1 [1 – i + 3i – 3]               &nb

...more

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abhishek gaurav

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Payal Gupta

Contributor-Level 10

27. Kindly go through the solution

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abhishek gaurav

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abhishek gaurav

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abhishek gaurav

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abhishek gaurav

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Payal Gupta

Contributor-Level 10

26. Kindly go through the solution

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abhishek gaurav

Contributor-Level 9

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abhishek gaurav

Contributor-Level 9

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abhishek gaurav

Contributor-Level 9

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Payal Gupta

Contributor-Level 10

25. Kindly go through the solution

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abhishek gaurav

Contributor-Level 9

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abhishek gaurav

Contributor-Level 9

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Payal Gupta

Contributor-Level 10

24. Kindly go through the solution

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Payal Gupta

Contributor-Level 10

23. x2 – x + 2 = 0

Comparing the given equation with ax2 + bx + c = 0

We have, a = 1, b = –1 andc = 2

Hence, discriminant of the equation is

b2 – 4ac = (-1)2 – 4 * 1 * 2 = 1 – 8 = –7

Therefore, the solution of the quadratic equation is

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