We have the following equation to solve
![E=I\cdot Z](https://img.qammunity.org/2023/formulas/mathematics/college/c45vhst81b1zgmpmzx8qeu8a5nx71wjls4.png)
Where E, I, and Z are complex numbers, therefore let's put it in numbers
![E=(4+4i)(7+3i)](https://img.qammunity.org/2023/formulas/mathematics/college/97x1kbkixznhrigyepbebcz64323t26ye3.png)
We can solve it directly into the rectangular form by doing the distrutive
Then
![(4+4i)(7+3i)=28+12i+28i+12i^2](https://img.qammunity.org/2023/formulas/mathematics/college/htxw980uumrb78xew37w7z8jlznrmrpxzi.png)
Remember that
![i^2=-1](https://img.qammunity.org/2023/formulas/mathematics/high-school/yq84fwq8mf651ezifio1x1gelf6zyjx7yj.png)
Then
![\begin{gathered} (4+4\imaginaryI)(7+3\imaginaryI)=28+12i+28i-12 \\ \\ (4+4\imaginaryI)(7+3\imaginaryI)=16+40i \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/bup1u6m5j64z3qc5nn556414g20oqbbkaa.png)
Now we have completely solved the problem!
![E=16+40i](https://img.qammunity.org/2023/formulas/mathematics/college/194mut91e92expb4up6ccl0dexhh5gh909.png)
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The second solution (usual)
When we have real engineering problems, we like to do multiplication and division with the polar form, then let's convert Z and I to the polar form
![\begin{gathered} I=4+4i=4√(2)\angle45° \\ \\ Z=7+3i=√(58)\angle23.2° \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/botqehiuhlpm36ertdy0yaks3qdvg22hwt.png)
Now to do the multiplication we multiple the magnitude and sum the phases (angles)
![\begin{gathered} ZI=4√(2)\cdot√(58)\angle45°+23.2° \\ \\ ZI=4√(116)\operatorname{\angle}68.2° \end{gathered}]()
We already have the result, now just put it in the rectangular form
![\begin{gathered} ZI=4√(116)\cdot\cos(68.2)+i4√(116)\sin(68.2) \\ \\ E=16+40i \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/eachxpee6c3k535b34kw4eflsroydih9yx.png)