• Jul 3rd 2008, 12:04 AM
sqleung

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Find the complex number $z$ that satisfies the equation:

$\frac {25}{z} - \frac {15}{\bar z} = 1 - 8i$

Given $\bar z$ is the conjugate of $z$ and $|z| = 2\sqrt{5}$

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This is what I've done so far:

Let $z = a + bi$
Therefore: $\bar z = a - bi$

$\sqrt {a^2 + b^2} = 2\sqrt{5}$

$a^2 + b^2 = 20$

$a^2 = 20 - b^2$

$a = \sqrt {20 - b^2}$

---

$\frac {25}{a + bi} - \frac {15}{a - bi} = 1 - 8i$

$\frac {25}{\sqrt {20 - b^2} + bi} - \frac {15}{\sqrt {20 - b^2} - bi} = 1 - 8i$

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Should I continue or should I stop before I start stuffing everything up? Is this correct? If not, could you please show me how to continue on?

Thank you - all help is of course, appreciated.
• Jul 3rd 2008, 12:12 AM
Moo
Hello,

Quote:

Originally Posted by sqleung

==================================

Find the complex number $z$ that satisfies the equation:

$\frac {25}{z} - \frac {15}{\bar z} = 1 - 8i$

Given $\bar z$ is the conjugate of $z$ and $|z| = 2\sqrt{5}$

==================================

This is what I've done so far:

Let $z = a + bi$
Therefore: $\bar z = a - bi$

$\sqrt {a^2 + b^2} = 2\sqrt{5}$

$a^2 + b^2 = 20$

$a^2 = 20 - b^2$

$a = \sqrt {20 - b^2}$

---

$\frac {25}{a + bi} - \frac {15}{a - bi} = 1 - 8i$

$\frac {25}{\sqrt {20 - b^2} + bi} - \frac {15}{\sqrt {20 - b^2} - bi} = 1 - 8i$

-------------------------------------------------

Should I continue or should I stop before I start stuffing everything up? Is this correct? If not, could you please show me how to continue on?

Thank you - all help is of course, appreciated.

Correct method, but I think you can shorten it a lot (Happy)

Let $z=a+ib$ and $\overline{z}=a-ib$.

$\frac{25}{z}-\frac{15}{\overline{z}}=1-8i$

We know that $z\overline{z}=|z|^2$. This is the key of the trick. You can check it, it's not difficult (Wink)

So the reflex will be to convert to a common denominator the LHS :

$\frac{25 \overline{z}-15 z}{z \overline{z}}=1-8i$

$z\overline{z}=|z|^2=(2 \sqrt{5})^2=20$

Thus, the equation is now :

$\frac{25 \overline{z}-15z}{20}=1-8i$

Substituting z and $\overline{z}$, and simplifying :

$\frac{10a-40ib}{20}=1-8i$

$\frac 12 a-2ib=1-8i$

Equate the real parts together, and the imaginary parts together and you will be done :)
• Jul 3rd 2008, 12:17 AM
sqleung