If $z = 1 + 2 i$ , then find the value of $z^{3} + 7 z^{2} - z + 16$ .

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Solution

We have,
$z = 1 + 2 i$
Since,
$z^{3} + 7 z^{2} - z + 16$
Put the value of $z$
Therefore,
$= {(1 + 2 i)}^{3} + {(1 + 2 i)}^{2} - (1 + 2 i) + 16$
$= 1 + 8 i^{3} + 6 i (1 + 2 i) + 1 + 4 i^{2} + 4 i - 1 - 2 i + 16$
$= 1 - 8 i + 6 i + 12 i^{2} + 1 - 4 + 2 i + 15$
$= 1 - 2 i - 12 - 3 + 2 i + 15$
$= 1$

Hence, the value is $1$ .

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If z=1+2i , then find the value of z3+7z2−z+16.

We have,z=1+2iSince,z3+7z2−z+16Put the value of zTherefore,=(1+2i)3+(1+2i)2−(1+2i)+16=1+8i3+6i(1+2i)+1+4i2+4i−1−2i+16=1−8i+6i+12i2+1−4+2i+15=1−2i−12−3+2i+15=1 Hence, the value is 1.

If $z = 1 + 2 i$ , then find the value of $z^{3} + 7 z^{2} - z + 16$ .