1.
\(z_1=\dfrac{i.\left(i^2\right)^{35}+\left(i^2\right)^{28}}{\left(i+1\right).\left[\left(i+1\right)^2\right]^7}=\dfrac{i.\left(-1\right)^{35}+\left(-1\right)^{28}}{\left(i+1\right)\left[i^2+2i+1\right]^7}\)
\(=\dfrac{-i+1}{\left(i+1\right)\left(2i\right)^7}=\dfrac{-i+1}{\left(i+1\right).2^7.i.\left(i^2\right)^3}=\dfrac{-i+1}{\left(i^2+i\right).2^7.\left(-1\right)^3}\)
\(=\dfrac{-i+1}{\left(-1+i\right).2^7.\left(-1\right)}=\dfrac{-i+1}{\left(-i+1\right).2^7}=\dfrac{1}{2^7}\)
b.
\(z_2=\left(\dfrac{1-\sqrt{3}i}{\sqrt{3}+i}\right)^{16}=\left(\dfrac{\left(1-\sqrt{3}i\right)\left(\sqrt{3}-i\right)}{\left(\sqrt{3}+i\right)\left(\sqrt{3}-i\right)}\right)^{16}\)
\(=\left(\dfrac{\sqrt{3}-4i+\sqrt{3}i^2}{3-i^2}\right)^{16}=\left(\dfrac{-4i}{4}\right)^{16}=\left(i\right)^{16}\)
\(=\left(i^2\right)^8=\left(-1\right)^8=1\)
2.
a.
\(\dfrac{z-2}{z+1}=3i+1\Leftrightarrow z-2=\left(3i+1\right)z+3i+1\)
\(\Leftrightarrow-3i.z=3i+3\)
\(\Leftrightarrow-i.z=i+1\)
\(\Leftrightarrow z=\dfrac{i+1}{-i}=\dfrac{i\left(i+1\right)}{-i^2}=i^2+i\)
\(\Leftrightarrow z=i-1\)
b.
\(z^4-z^2-2=0\Leftrightarrow\left(z^2+1\right)\left(z^2-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}z^2=-1\\z^2=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}z=\pm i\\z=\pm\sqrt{2}\end{matrix}\right.\)
giúp e câu 55,56,57,58,59 ạ. Đề ghi khó hiểu quá ạ
