Question

Calcule:

(c) i(1 + i)|2 − 3i| + i


(D) $(3-4i) ^ 2 + i \frac |3-4i|$

Answer 1

Unidade imaginária dos números complexos:

$\boxed{\boxed{i^{2}=-1}}$

Forma algébrica de um número complexo:

$\boxed{\boxed{z=x+yi}}\\\\onde:\\\\x=Re(z)~~(parte~real~de~z)\\y=Im(z)~~(parte~imagin\'aria~de~z)$

Definição do módulo de um número complexo:

$|z|=\sqrt{a^{2}+b^{2}}=\sqrt{(Re(z))^{2}+(Im(z))^{2}}$

Definição de conjugado de um número complexo:

$z=a+bi~~\rightarrow~~\boxed{\boxed{\overline{z}=a-bi}}$
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C)

$i\cdot(1+i)\cdot|2-3i|+i=(i+i^{2})\cdot\sqrt{2^{2}+(-3)^{2}}+i\\\\i\cdot(1+i)\cdot|2-3i|+i=(i+[-1])\cdot\sqrt{4+9}+i\\\\i\cdot(1+i)\cdot|2-3i|+i=(i-1)\cdot\sqrt{13}+i\\\\i\cdot(1+i)\cdot|2-3i|+i=i\sqrt{13}-\sqrt{13}+i\\\\i\cdot(1+i)\cdot|2-3i|+i=-\sqrt{13}+i\sqrt{13}+i\\\\\boxed{\boxed{i\cdot(1+i)\cdot|2-3i|+i=-\sqrt{13}+(\sqrt{13}+1)i}}$

D)

$\dfrac{(\overline{3-4i})^{2}+i}{|3-4i|}=\dfrac{(3^{2}+2\cdot3\cdot4i+[4i]^{2})+i}{\sqrt{3^{2}+(-4)^{2}}}\\\\\\\dfrac{(\overline{3-4i})^{2}+i}{|3-4i|}=\dfrac{(9+24i+16i^{2})+i}{\sqrt{9+16}}\\\\\\\dfrac{(\overline{3-4i})^{2}+i}{|3-4i|}=\dfrac{9+24i+16(-1)+i}{\sqrt{25}}\\\\\\\dfrac{(\overline{3-4i})^{2}+i}{|3-4i|}=\dfrac{9-23i-16}{5}\\\\\\\dfrac{(\overline{3-4i})^{2}+i}{|3-4i|}=\dfrac{-7+25i}{5}\\\\\\\dfrac{(\overline{3-4i})^{2}+i}{|3-4i|}=-\dfrac{7}{5}-\dfrac{25}{5}i$

$\boxed{\boxed{\dfrac{(\overline{3-4i})^{2}+i}{|3-4i|}=-\dfrac{7}{5}+5i}}$