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Question

If z=2−3i, show that z2−4z+13=0 and hence find the value of 4z3−3z2+169.

A
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B
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C
2
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3
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Solution

The correct option is A 0
z=23i
(z2)=3i
(z2)2=(3i)2
z24z+4=9i2
z24z+13=0
Therefore,
4z33z2+169=4z(z24z+13)+13(z24z+13)=4z(0)+13(0)=0

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