If x = 3 + i , then x^3 - 3x^2

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4-1.Complex numbers

easy

If $x = 3 + i$, then $x^3 - 3x^2 - 8x + 15 = $

$6$

$10$

$-18$

$-15$

Solution

(d) $x = 3 + i\, \Rightarrow \,\,x – 3 = i\,$
$ \Rightarrow $ $\,x^2 – 6x + 10 = 0$ ==> $x^3 – 3x^2 – 8x + 15$
$ = x(x^2 – 6x + 10) + 3\,(x^2 – 6x + 10) – 15$
$ = x(0) + 3(0) – 15 = – 15$.

Standard 11

Mathematics

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

hard

For any two complex numbers $z_{1}$ and $z_{2},$ prove that

$\operatorname{Re}\left(z_{1} z_{2}\right)=\operatorname{Re} z_{1} \operatorname{Re} z_{2}-\operatorname{Im} z_{1} \operatorname{Im} z_{2}$

medium

If $4 x+i(3 x-y)=3+i(-6),$ where $x$ and $y$ are real numbers, then find the values of $x$ and $y.$

easy

Solve the equation $x^{2}+\frac{x}{\sqrt{2}}+1=0$

medium

If $(1 – i)^n = 2^n,$ then $n = $

medium

If $x = 3 + i$, then $x^3 - 3x^2 - 8x + 15 = $

Solution

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For any two complex numbers $z_{1}$ and $z_{2},$ prove that

$\operatorname{Re}\left(z_{1} z_{2}\right)=\operatorname{Re} z_{1} \operatorname{Re} z_{2}-\operatorname{Im} z_{1} \operatorname{Im} z_{2}$