=1+23+62i=33(1+22i)=1+22i i) i−35=i351=(i2)17i1=−i1×ii=−i2i=i of the complex number given in the Exercise 2. i9+i19 3. i−39

Name: Video solution 1: =1+23+62i=33(1+22i)=1+22i i) i−35=i351=(i2)17i1=−i1×ii=−i2i=i of the complex number given in the Exercise 2. i9+i19 3. i−39
Uploaded: 2022-09-14T12:08:21.000Z
Duration: 3 min
Description: Video solution 1: =1+23+62i=33(1+22i)=1+22i i) i−35=i351=(i2)17i1=−i1×ii=−i2i=i of the complex number given in the Exercise 2. i9+i19 3. i−39

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={3+6 {2} i}{1+2}={3(1+2 {2} i)}{3}=1+2 {2} i i) i^{-35}={1}{i

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=1+23+62i=33(1+22i)=1+22i
i) i−35=i351=(i2)17i1=−i1×ii=−i2i=

$= \frac{3 + 6 2 i}{1 + 2} = \frac{3 ( 1 + 2 2 i )}{3} = 1 + 22 i$ i) $i^{- 35} = \frac{1}{i ^{35}} = \frac{1}{( i ^{2} ) ^{17} i} = \frac{1}{- i} \times \frac{i}{i} = \frac{i}{- i ^{2}} = i$ of the complex number given in the Exercise 2. $i^{9} + i^{19}$ 3. $i^{- 39}$

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Updated on: Sep 14, 2022

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= \frac{3 + 6 2 i}{1 + 2} = \frac{3 ( 1 + 2 2 i )}{3} = 1 + 22 i

i^{- 35} = \frac{1}{i ^{35}} = \frac{1}{( i ^{2} ) ^{17} i} = \frac{1}{- i} \times \frac{i}{i} = \frac{i}{- i ^{2}} = i

of the complex number given in the Exercise 2.

i^{9} + i^{19}

i^{- 39}

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Question Text	$= \frac{3 + 6 2 i}{1 + 2} = \frac{3 ( 1 + 2 2 i )}{3} = 1 + 22 i$ i) $i^{- 35} = \frac{1}{i ^{35}} = \frac{1}{( i ^{2} ) ^{17} i} = \frac{1}{- i} \times \frac{i}{i} = \frac{i}{- i ^{2}} = i$ of the complex number given in the Exercise 2. $i^{9} + i^{19}$ 3. $i^{- 39}$
Updated On	Sep 14, 2022
Topic	Vector and 3D
Subject	Mathematics
Class	Class 11
Answer Type	Video solution: 1
Upvotes	119
Avg. Video Duration	3 min

$= \frac{3 + 6 2 i}{1 + 2} = \frac{3 ( 1 + 2 2 i )}{3} = 1 + 22 i$ i) $i^{- 35} = \frac{1}{i ^{35}} = \frac{1}{( i ^{2} ) ^{17} i} = \frac{1}{- i} \times \frac{i}{i} = \frac{i}{- i ^{2}} = i$ of the complex number given in the Exercise 2. $i^{9} + i^{19}$ 3. $i^{- 39}$

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=1+23+62​i​=33(1+22​i)​=1+22​i i) i−35=i351​=(i2)17i1​=−i1​×ii​=−i2i​=i of the complex number given in the Exercise 2. i9+i19 3. i−39

Filo tutor solutions (1)

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Students who ask this question also asked

$= \frac{3 + 6 2 i}{1 + 2} = \frac{3 ( 1 + 2 2 i )}{3} = 1 + 22 i$ i) $i^{- 35} = \frac{1}{i ^{35}} = \frac{1}{( i ^{2} ) ^{17} i} = \frac{1}{- i} \times \frac{i}{i} = \frac{i}{- i ^{2}} = i$ of the complex number given in the Exercise 2. $i^{9} + i^{19}$ 3. $i^{- 39}$