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# complex numbers

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Determine the complex number z satisfying the equation $2z - 3i \overline{z} = -7 + 2i$. Note that $\overline{z}$ denotes the conjugate of z.

Apr 22, 2021

#1
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Let a and b be real numbers, such that a is the real part of z, and b is the imaginary part. Therefore,

$$2(a+bi)-3i(a-bi)=7-2i\\ 2a+2bi-3ai-3b=7-2i\\ (2a-3b)+(2b-3a)i=7-2i$$

Notice the real part of the right-hand side is 2a-3b, and the real part of the left-hand side is 7, so $$2a-3b=7$$. Similarly, the imaginary part of the right-hand side is 2b-3a, and the imaginary part of the left-hand side is -2, so that means that $$-3a+2b=-2$$. Now it's just a system of 2 linear equations, and can be easily solved with elimination:

$$6a-9b=21\\-6a+4b=-4\\-5b=17\\b=-\frac{17}{5}\\-3a+2(-\frac{17}{5})=-2\\-3a-\frac{34}{5}=-2\\ -3a=\frac{24}{5}\\a=-\frac{8}{5}$$

Therefore, the complex number z is equal to $$\boxed{-\frac{8}{5}-\frac{17}{5}i}$$

Apr 22, 2021

#1
+2
$$2(a+bi)-3i(a-bi)=7-2i\\ 2a+2bi-3ai-3b=7-2i\\ (2a-3b)+(2b-3a)i=7-2i$$
Notice the real part of the right-hand side is 2a-3b, and the real part of the left-hand side is 7, so $$2a-3b=7$$. Similarly, the imaginary part of the right-hand side is 2b-3a, and the imaginary part of the left-hand side is -2, so that means that $$-3a+2b=-2$$. Now it's just a system of 2 linear equations, and can be easily solved with elimination:
$$6a-9b=21\\-6a+4b=-4\\-5b=17\\b=-\frac{17}{5}\\-3a+2(-\frac{17}{5})=-2\\-3a-\frac{34}{5}=-2\\ -3a=\frac{24}{5}\\a=-\frac{8}{5}$$
Therefore, the complex number z is equal to $$\boxed{-\frac{8}{5}-\frac{17}{5}i}$$