
How do you graphically divide complex numbers?
Answer
541.2k+ views
Hint: These types of problems are pretty straight forward and are very simple to solve. We can go through the problem very quickly once we understand the underlying concepts of complex numbers. Complex numbers are generally represented as \[a+ib\] , here ‘a’ is the real part and ‘b’ is the complex part. The ‘i’ in this equation is known as iota and is generally used to represent complex numbers. Complex numbers can also be represented graphically, just like we can plot real numbers and equations in the coordinate plane, we can plot complex numbers in the argand plane. The argand plane consists of a real horizontal axis and an imaginary vertical axis. Any complex number in its polar form is represented by its distance from the origin and an angle which it makes with the real axis, known as argument. It is represented as \[r{{e}^{i\theta }}\] .
Complete step-by-step solution:
Now we start off with the solution to the problem by assuming two complex numbers, \[{{z}_{1}}={{r}_{1}}{{e}^{i{{\theta }_{1}}}}\] and \[{{z}_{2}}={{r}_{2}}{{e}^{i{{\theta }_{2}}}}\] . Now we divide these two complex numbers and we write,
\[\dfrac{{{z}_{1}}}{{{z}_{2}}}=\dfrac{{{r}_{1}}{{e}^{i{{\theta }_{1}}}}}{{{r}_{2}}{{e}^{i{{\theta }_{2}}}}}\] . Now we bring the denominator of the right hand side of the equation to the numerator and we write,
\[\Rightarrow \dfrac{{{z}_{1}}}{{{z}_{2}}}=\dfrac{{{r}_{1}}}{{{r}_{2}}}{{e}^{i{{\theta }_{1}}}}\times {{e}^{-i{{\theta }_{2}}}}\]
We now perform subtraction of the powers in the right hand side of the equation to get,
\[\Rightarrow \dfrac{{{z}_{1}}}{{{z}_{2}}}=\dfrac{{{r}_{1}}}{{{r}_{2}}}{{e}^{i{{\theta }_{1}}-i{{\theta }_{2}}}}\]
Taking ‘i’ common on the power we get,
\[\Rightarrow \dfrac{{{z}_{1}}}{{{z}_{2}}}=\dfrac{{{r}_{1}}}{{{r}_{2}}}{{e}^{i\left( {{\theta }_{1}}-{{\theta }_{2}} \right)}}\]
Thus we can clearly see that when we divide two complex numbers, the resultant complex number formed has argument as the difference of the two given complex numbers, i.e.\[{{\theta }_{1}}-{{\theta }_{2}}\] .
Note: Solving these problems is a pretty easy task. We must thus remember that when we multiply two complex numbers the resultant complex number formed has an argument as the sum of the arguments of the two given complex numbers. When we divide two complex numbers the resultant complex number formed has an argument as the difference of the arguments of the two given complex numbers. The magnitude of the resultant complex number however gets divided.
Complete step-by-step solution:
Now we start off with the solution to the problem by assuming two complex numbers, \[{{z}_{1}}={{r}_{1}}{{e}^{i{{\theta }_{1}}}}\] and \[{{z}_{2}}={{r}_{2}}{{e}^{i{{\theta }_{2}}}}\] . Now we divide these two complex numbers and we write,
\[\dfrac{{{z}_{1}}}{{{z}_{2}}}=\dfrac{{{r}_{1}}{{e}^{i{{\theta }_{1}}}}}{{{r}_{2}}{{e}^{i{{\theta }_{2}}}}}\] . Now we bring the denominator of the right hand side of the equation to the numerator and we write,
\[\Rightarrow \dfrac{{{z}_{1}}}{{{z}_{2}}}=\dfrac{{{r}_{1}}}{{{r}_{2}}}{{e}^{i{{\theta }_{1}}}}\times {{e}^{-i{{\theta }_{2}}}}\]
We now perform subtraction of the powers in the right hand side of the equation to get,
\[\Rightarrow \dfrac{{{z}_{1}}}{{{z}_{2}}}=\dfrac{{{r}_{1}}}{{{r}_{2}}}{{e}^{i{{\theta }_{1}}-i{{\theta }_{2}}}}\]
Taking ‘i’ common on the power we get,
\[\Rightarrow \dfrac{{{z}_{1}}}{{{z}_{2}}}=\dfrac{{{r}_{1}}}{{{r}_{2}}}{{e}^{i\left( {{\theta }_{1}}-{{\theta }_{2}} \right)}}\]
Thus we can clearly see that when we divide two complex numbers, the resultant complex number formed has argument as the difference of the two given complex numbers, i.e.\[{{\theta }_{1}}-{{\theta }_{2}}\] .
Note: Solving these problems is a pretty easy task. We must thus remember that when we multiply two complex numbers the resultant complex number formed has an argument as the sum of the arguments of the two given complex numbers. When we divide two complex numbers the resultant complex number formed has an argument as the difference of the arguments of the two given complex numbers. The magnitude of the resultant complex number however gets divided.
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