Ncert Solutions Maths class 11th

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Payal Gupta

Contributor-Level 10

15. Kindly go through the solution

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Payal Gupta

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14. Kindly go through the solution

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Payal Gupta

Contributor-Level 10

12. Let z = -i

Then z¯ = i

And |z|2 = (-1)2 = 1

So, multiplicative inverse of –i is given by

z-1 = z¯|z| = i1 = I = 0 + i1

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Payal Gupta

Contributor-Level 10

11. Let Z = 4 – 3i

Then z¯ = 4 + 3i

And, z|2 = 42+ (-3)2= 16 + 9 = 25

Hence, z-1 = z¯|z|2 = 4+3i25 = 425 + i325

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Payal Gupta

Contributor-Level 10

10.(213i)3

[(2+13i)]3

(1)3[(2+13i)3]

(8)[8+i327+3.2.13i(2+i3)] [since, (a + b)3 = a3 + b3 + 3ab(a + b)]

= (–1) [8i27+(2i*2)+(2i*i3)] [since, i3 = i2.i = –i and i2 = –1]

= (–1) [8i27+4i+2i23]

= (–1) [823+i(4127)]

223 i10727

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Payal Gupta

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7.  [ (13+i73)+ (4+i13)] (43+i)

13 + i73 + 4 + i13 + 43 – i

(13+4+43)+i (73+131)

1+12+43 + i7+133

173 + i53

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Payal Gupta

Contributor-Level 10

6.  (15+i25) –  (4+i52)

15 + i25 – 4 – i52

15 – 4 + i  (2552)

1205 + i  (42510)

195 + i  (2110)

195 – i 2110

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Payal Gupta

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5. (1 – i) – (–1 + i6)

= 1 – I + 1 – i6

= 2 – 7i

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Payal Gupta

Contributor-Level 10

4. 3 (7 + i7) + I (7 + i7)

= 21 + 21i + 7i + 7i2

= 21 + 28i + 7 (–1) [since i2 = –1]

= 21 + 28i – 7

= 14 + 28i

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Payal Gupta

Contributor-Level 10

1. (5i) -35i

= 5  (35) *i2          [since i2 = –1]

= –3 (–1)

= 3

So, (5i)  (35i) = 3 + i0

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