# Complex Number

• Jun 18th 2009, 01:29 AM
cloud5
Complex Number
Can help me to solve this?

1. Given that $(1+5i)p-2q=3+7i$, 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 $zz*+2zi=12+6i$, find the possible values of z.

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

1. p=2-i, q=2+i
2. 3-i, 3+3i
• Jun 18th 2009, 02:01 AM
yeongil
Quote:

Originally Posted by cloud5
3. Given that z=x+yi and $w= \frac {z+8i}{z-6}$. If w is totally imaginary, show that $x^2+y^2+2x-48=0$.

We solved a problem like this one in this thread: http://www.mathhelpforum.com/math-he...ex-number.html

01
• Jun 18th 2009, 02:07 AM
yeongil
Going a bit backwards here:

Quote:

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

$zz*+2zi=12+6i$

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

\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:
\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.

Quote:

Originally Posted by cloud5
1. Given that $(1+5i)p-2q=3+7i$, 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.

01