# Find the complex number z such that 5iz+3z* +16 = 8i. Give your answer in the form a + bi, where a and b are real numbers.

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Firstly, we note that z being a complex number can be expressed in the form z = a + bi. If we then take the complex conjugate of this expression, the real numbers remain the same (as they are their own c conjugates) but the c conjugate of i is -i, therefore z* = a - bi. We then insert these expressions into the given equation, so

5 i (a + bi)  + 3 (a - bi) + 16 = 8 i

We then expand the brackets and rearrange, remebering that i2 = -1, such that

5 i a - 5 b + 3 a - 3 b i + 16 = 8i.

We can then split the equation up into real and complex parts (i.e the terms that are not functions of i and are functions of i respectively) and treat each equation seperately, so we have

-5 b + 3 a + 16 = 0,       5a - 3b - 8 = 0

This is a simple simultaneous equation to solve, and we this find that

16b = 104, 16a = 88

so z is thus given by z = 11/2 + 13/2 i

QED

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