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Question

Let Z1,Z2C such that Z21+Z22R. If Z1(Z213Z22)=10 and Z2(3Z21Z22)=30, Find the value of (Z21+Z22).

A
Z21+Z22=10
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B
Z21+Z22=15
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C
Z21+Z22=20
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D
Z21+Z22=50
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Solution

The correct option is A Z21+Z22=10
Z313Z!Z22=10 ...(1)
3Z2Z21Z32=30
3Z2Z21iZ32i=Z32i=30i .... (2)
eqn.(1)+(2)=
Z313Z1Z22+3Z2Z21iz32i=10+30i
(Z1+iZ2)3=10+30i
|(Z1+iZ2)3)|=|10+30i|
|Z1+iZ2|3=100+900
(Z21+Z22)3=1000
Z21+Z22=(1000)1/3
Z21+Z22=10

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