Question

If $z=3-4i$, then $z^4-3z^3+3z^2+99z-95$ is equal to

• A. $5$
• B. $6$
• C. $-5$
• D. $-4$

Answer 1

The correct option is A $5$

The given polynomial $z^4-3z^3+3z^2+99z-95$ can be simplified as follows:

$z^4-3z^3+3z^2+99z-95=z^4-3z^3+3z^2-z+100z-95=z(z^3-3z^2+3z-1+100)-95=z\left[\right(z^3-3z^2+3z-1)+100]-95=z\left[\right(z-1{)}^3+100]-95(\because (a-b{)}^3=a^3-b^3-3a^{2}b+3a{b}^2)$

Since $z=3-4i$, therefore, we have:

$z\left[\right(z-1{)}^3+100]-95=(3-4i\left)\right[(3-4i-1{)}^3+100]-95=(3-4i)\left[\right(2-4i{)}^3+100]-95=(3-4i\left)\right[\left(2\right(1-2i){)}^3+100]-95=(3-4i)\left[8\right(1^3-(2i{)}^3-(3\times 1^2\times 2i)+(3\times 1\times \left(2i{)}^2\right)+100]-95$

$(\because (a-b{)}^3=a^3-b^3-3a^{2}b+3a{b}^2)=(3-4i)\left[8\right(1-8i^3-6i+12i^2)+100]-95(\because i^2=-1)=(3-4i)\left[8\right(1-8i-6i-12)+100]-95=(3-4i)\left[8\right(2i-11)+100]-95=(3-4i)(16i-88+100)-95=(3-4i)(16i+12)-95$

$=4(3-4i)(3+4i)-95=4(3^2-(4i{)}^2)-95(\because a^2-b^2=(a+b\left)\right(a-b\left)\right)=4(9-16i^2)-95=4(9+16)-95(\because i^2=-1)=(4\times 25)-95=100-95=5$

Hence, $z^4-3z^3+3z^2+99z-95=5$, if $z=3-4i$.