1. ## Complex Number

Can help me to solve this?

1. Given that $\displaystyle (1+5i)p-2q=3+7i$, find the values of p and q when p and q are respectively a complex number and its conjugate.

2. Given that the complex number z and its conjugate z* satisfy the equation $\displaystyle zz*+2zi=12+6i$, find the possible values of z.

3. Given that z=x+yi and $\displaystyle w= \frac {z+8i}{z-6}$. If w is totally imaginary, show that $\displaystyle x^2+y^2+2x-48=0$.

1. p=2-i, q=2+i
2. 3-i, 3+3i

2. Originally Posted by cloud5
3. Given that z=x+yi and $\displaystyle w= \frac {z+8i}{z-6}$. If w is totally imaginary, show that $\displaystyle x^2+y^2+2x-48=0$.
We solved a problem like this one in this thread: http://www.mathhelpforum.com/math-he...ex-number.html

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3. Going a bit backwards here:

Originally Posted by cloud5
2. Given that the complex number z and its conjugate z* satisfy the equation $\displaystyle zz*+2zi=12+6i$, find the possible values of z.
$\displaystyle zz*+2zi=12+6i$

If $\displaystyle z = a + bi$, then $\displaystyle z* = a - bi$:

\displaystyle \begin{aligned} (a + bi)(a - bi) + 2(a + bi)i &= 12+6i \\ a^2 + b^2 + 2ai -2b &= 12 + 6i \\ (a^2 + b^2 - 2b) + 2ai &= 12 + 6i \end{aligned}

Now equate the real and imaginary coefficients:
\displaystyle \begin{aligned} a^2 + b^2 - 2b &= 12 \\ 2a &= 6 \\ a &= 3 \\ 9 + b^2 - 2b &= 12 \\ b^2 - 2b -3 &= 0 \\ (b - 3)(b + 1) &= 0 \\ b &= 3\;\;{\color{red}or} \\ b &= -1 \end{aligned}

So the answers, in the form of a + bi, are
3 + 3i and
3 - i.

Originally Posted by cloud5
1. Given that $\displaystyle (1+5i)p-2q=3+7i$, find the values of p and q when p and q are respectively a complex number and its conjugate.
This looks like the same problem as #2. Substitute a + bi for p and a - bi for q. Set the real and imaginary coefficients equal to each other and solve for a and b.

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