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Question

The solution of the equation |z|z=1+2i is

A
322i
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B
32+2i
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C
232i
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D
None of these
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Solution

The correct option is A 322i
We have |z|z=1+2i

x2+y2(x+iy)=1+2i where z=x+iy

x2+y2x=1 and y=2

[Comparing real and imaginary parts]

x=32 and y=2

The solution of the given equation is 322i.

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