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Question

If |z|z=1+2i then z=

A
232i
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B
322i
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C
3+4i
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D
34i
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Solution

The correct option is C 322i
Given,
|z|z=1+2i
Let z=x+iy

So, |z|z=1+2i

(x2+y2x)iy=1+2i

y=2(1)

So, x2+y2x=1

x2+4x=1

x2+4=(1+x)2=x2+2x+1

2x=3

x=32(2)

Hence,
z=322i

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