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Q.

How many number are in the form z=a+ib, when a and bare real numbers such that  z.z¯=25and a+b=7

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a

1

b

2

c

3

d

4

answer is B.

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Detailed Solution

Given z=a+ib and zz¯=25

(a+ib)(aib)=25

a2+b2=25;a+b=7

a=7b

(7b)2+b2=25

2b214b+24=0

b27b+12=0

(b3)(b4)=0

b=3 or b=4

a=4 or a=3

z=4+3i or z=3+4i

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