Question

find multiplactive inverse of z=(2+3i)²​

Answer 1

Step-by-step explanation:

Step-by-step explanation:

Given,

z = {(2 + \sqrt{3}i) }^{2}z=(2+

3

i)

2

= > z = {(2)}^{2} + {( \sqrt{3}i) }^{2} + 2.2. (\sqrt{3}) i=>z=(2)

2

+(

3

i)

2

+2.2.(

3

)i

z = 4 - 3 + (4 \sqrt{3}) iz=4−3+(4

3

)i

= > z = 1 + (4 \sqrt{3}) i=>z=1+(4

3

)i

So,

{z}^{ - 1} = \frac{1}{1 + 4 \sqrt{3} i }z

−1

=

1+4

3

i

1

= > {z}^{ - 1} = \frac{1 - 4 \sqrt{3}i }{(1 + 4 \sqrt{3}i)(1 - 4 \sqrt{3} i) }=>z

−1

=

(1+4

3

i)(1−4

3

i)

1−4

3

i

= > {z}^{ - 1} = \frac{1 - 4 \sqrt{3} i}{1 + 48}=>z

−1

=

1+48

1−4

3

i

= > {z}^{ - 1} = \frac{1}{49} - \frac{4 \sqrt{3} }{49} i=>z

−1

=

49

1

−

49

4

3

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