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Q.

If z=−53+4i5−7i3−4i68+7i5+7i8−7i9 then z is

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a

purely real

b

purely imaginary

c

a+ib, where a≠0,b≠0

d

a+ib, where b=4

answer is A.

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Detailed Solution

z=−53+4i5−7i3−4i68+7i5+7i8−7i9⇒ z¯=−53−4i5+7i3+4i68−7i5−7i8+7i9=−53+4i5−7i3−4i68+7i5+7i8−7i9=z(Taking transpose)Hence, z is purely real.
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If z=−53+4i5−7i3−4i68+7i5+7i8−7i9 then z is