Question

If Z1=2+3i and Z2=4-5i then find a)Z1+Z2b)Z1÷Z2c)Z1-Z2d)Z1×Z2e)Z1bar ×Z2bar​

Answer 1

Step-by-step explanation:

a).

• Z1+Z2
• 2+3i+4-5i
• 2+4+3i-5i
• 6-2i

b).

• Z1÷Z2
• 2+3i/4-5i
• 2+3i=4-5i
• 2-4=-5i-3i
• -2=-8i
• I=-2/-8
• I=1/4

c).

• Z1-Z2
• (2+3i)-(4-5i
• 2+3i-4+5i
• -2+8i

d).

• Z1×Z2
• (2+3i)(4-5i)
• 2(4-5i)+3i(4-5i)
• 8-10i+12i-15i²
• 8+2i+15i²

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Answer 2

Answer:

Required value of $z_1 + z_2$ is 7-2i.

and required value of $z_1 \div z_2$ is $\frac{7 - 22i}{9}$

and required value of $z_1 - z_2$ is (-2+8i)

and required value of $z_1 \times z_2$ is (22+2i)

and required value of Z1bar ×Z2bar is 22-7i

Step-by-step explanation:

Given,

$z_1 = 2 + 3i \\ z_2 = 4 - 5i$

Now,

$z_1 + z_2 \\ = 2 + 3i + 4 - 5i \\ = (3 + 4) + i(3 - 5) \\ = 7 - 2i$

and

$z_1 \div z_2 \\ = \frac{2 + 3i}{4 - 5i} \\ = \frac{(2 + 3i)(4 + 5i)}{(4 - 5i)(4 + 5i)} \\ = \frac{8 + 10i + 12i - 15}{ {4}^{2} - {(5i)}^{2} } \\ = \frac{ - 7 + 22i}{16 - 25} \\ = \frac{7 - 22i}{9}$

and

$z_1 - z_2 = 2 + 3i - (4 - 5i) \\ = 2 + 3i - 4 + 5i \\ = - 2 + 8i$

and

$z_1 \times z_2 = (2 + 3i) \times (4 - 5i) \\ = 8 - 10i + 12i - 15 {i}^{2} \\ = 8 + 2i + 15 \\ = 22 + 2i$

and Z1 bar ×Z2 bar

$= (2 - 3i) \times (4 + 5i) \\ = 8 + 5i - 12i - 15 {i}^{2} \\ = 8 - 7i + 15 \\ = 22 - 7i$

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