Let
"z = x + iy \\Rightarrow z* = x - iy"
Then
"\\begin{array}{l}\n2(x + iy) - i(x - iy) = 3(3 - 5i)\\\\\n2x + 2iy - ix + {i^2}y = 9 - 15i\\\\\n2x - y + i(2y - x) = 9 - 15i\n\\end{array}"
Then
"\\left\\{ \\begin{array}{l}\n2x - y = 9\\\\\n2y - x = - 15\n\\end{array} \\right."
"\\left\\{ \\begin{array}{l}\n2x - y = 9\\\\\n3x = 3\n\\end{array} \\right."
"\\left\\{ \\begin{array}{l}\nx = 1\\\\\ny = -7\n\\end{array} \\right."
Answer: "z = 1 - 7i"
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