Question

conjucate of complex number (3-i) ²/2+i
​

Answer 1

Answer:

Answer

Let z=−2+2

3

i

Then, modulus of z = ∣z∣=

(−2)

2

+(2

3

)

2

=

16

=4.

Answer 2

Step-by-step explanation:

is Given That ,

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

2+i

(3−i)

2

now By using Identity,

{(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab(a−b)

2

=a

2

+b

2

−2ab

We Get,

\frac{ ({3}^{2} + {i}^{2} - 2 \times 3 \times i )}{2 + i}

2+i

(3

2

+i

2

−2×3×i)

We Know that,

{i}^{2} = - 1i

2

=−1

putting value,

we get,

\frac{9 - 1 - 6i}{2 + i}

2+i

9−1−6i

z = \frac{8 - 6i}{2 + i}z=

2+i

8−6i

Now , Rationalize it,

\frac{8 - 6i}{2 + i} \times \frac{2 - i}{2 - i}

2+i

8−6i

×

2−i

2−i

z = \frac{(8 - 6i)(2 - i)}{(2 + i)(2 - i)}z=

(2+i)(2−i)

(8−6i)(2−i)

z = \frac{16 - 8i - 12i + 6 {i}^{2} }{ {2}^{2} - {i}^{2} }z=

2

2

−i

2

16−8i−12i+6i

2

Again Filling The Value of

{i}^{2} = - 1i

2

=−1

We get,

z = \frac{16 -20i - 6}{4 - ( - 1)}z=

4−(−1)

16−20i−6

z = \frac{10 - 20i}{4 + 1}z=

4+1

10−20i

z = \frac{10 - 20i}{5}z=

5

10−20i

Split Both Imaginary and Real Roots,

z = \frac{10}{5} - \frac{20i}{5}z=

5

10

−

5

20i

z = 2 - 4iz=2−4i

Now We Have To Find the Conjugate ,

Conjugate Means To Simply Change the Signs of imaginary Root,

Conjugate is equal To ,

\frac{}{z} = 2 + 4i

z

=2+4i