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suter [353]
4 years ago
9

E=IZ Where E=(5+9i) and I=(8+7i) How do I solve for Z?

Mathematics
1 answer:
SashulF [63]4 years ago
6 0

Answer:

Z =  \frac{37i }{113 }  -  \frac{23}{113}

Step-by-step explanation:

We have the complex number relationship as:

E=IZ

Where E=(5+9i) and I=(8+7i)

We solve for Z to get;

Z=E/L

We then substitute the expressions to get:

Z =  \frac{5 + 9i}{8 + 7i}

We rationalize to get:

Z =  \frac{5 + 9i}{8 + 7i}  \times  \frac{8 - 7i}{8  - 7i}

We simplify to obtain:

Z =  \frac{40 - 35i + 72i - 63}{ {8}^{2} +  {7}^{2} }

This gives

Z =  \frac{37i - 23}{113 }

Or

Z =  \frac{37i }{113 }  -  \frac{23}{113}

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