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]]>-b) Find the three different roots \(u_1\), \(u_2\) or \(u_3\) of \(z\), where. \(u_1^3 = u_2^3 = u_3^3 = z\), but \(u_1 \neq u_2\), \(u_2 \neq u_3\), \(u_3 \neq u_1\). -
-
-Answers: Give all answers exactly! The answers in b) must also be given in polar form.
- The easiest way to enter the exponential function is as exp().
-
-Example: \(\dfrac{6}{5}\, \exp\left(\dfrac{\mathrm i\, \pi}{7}\right) = \dfrac{6}{5}\, e^{\frac{\mathrm i\, \pi}{7}}\) is entered as (6/5)*exp(%i*%pi/7).
-
| a) | -\(z =~\)[[input:SAns1]]\(\, \exp(\)[[input:SAns2]]\(\, \mathrm{i})\) | -|
| - | [[validation:SAns1]] | -[[validation:SAns2]] | -
| b) | -\(u_1 =~\)[[input:SAns3]] | -[[validation:SAns3]] [[feedback:prt3]] | -
| - | \(u_2 =~\)[[input:SAns4]] | -[[validation:SAns4]] [[feedback:prt4]] | -
| - | \(u_3 =~\)[[input:SAns5]] | -[[validation:SAns5]][[feedback:prt5]] | -
| a) | +\(z =~\)[[input:SAns1]]\(\, \exp(\)[[input:SAns2]]\(\, \mathrm{i})\) | +|
| + | [[validation:SAns1]] | +[[validation:SAns2]] | +
| b) | +\(u_1 =~\)[[input:SAns3]] | +[[validation:SAns3]] [[feedback:prt3]] | +
| + | \(u_2 =~\)[[input:SAns4]] | +[[validation:SAns4]] [[feedback:prt4]] | +
| + | \(u_3 =~\)[[input:SAns5]] | +[[validation:SAns5]][[feedback:prt5]] | +
The argument \(\theta\) is given by the angle between the positive real axis and the vector from the origin to \(z\), with the requirement \(-\pi < \theta \leq \pi\). -We can immediately see from the sign of \(z\) that \(\theta = \displaystyle {@arg_z@}\), since the real part is \(0\).
-Now we can write \(z\) in polar form: \[z = {@radius@} \, e^{ {@arg_z@} \, \mathrm i }\]
- -b) Using the following formula, we can then calculate the complex \(n\):th roots:
-\[\sqrt[n]{z} = \sqrt[n]{r} \,\exp\left( \mathrm i\,\frac{\theta + 2k\pi}{n}\right), \quad k = 0, 1, \dots , n-1.\]
- -With \(k=0\) we get -\[\begin{split} u_1 &= \sqrt[3]{{@radius@}} \exp\left( \mathrm i\, \frac{{@arg_z@}}{3} \right) \\ -&= {@polarform_simp(u1)@}. -\end{split}\]
- -Similarly with \(k=1\) we get -\[\begin{split} u_2 &= \sqrt[3]{{@radius@}} \exp \left( \mathrm i\, \frac{{@arg_z@} + 2\pi}{3}\right) \\ -&= {@polarform_simp(u2)@}, -\end{split}\] -
-and with \(k=2\):
-\[\begin{split} u_3 &= \sqrt[3]{{@radius@}}\exp \left( \mathrm i \, \frac{{@arg_z@} + 4\pi}{3}\right) \\
-&= {@polarform_simp(u3)@}.
+
The argument \(\theta\) is given by the angle between the positive real axis and the vector from the origin to \(z\), with the requirement \(-\pi < \theta \leq \pi\). +We can immediately see from the sign of \(z\) that \(\theta = \displaystyle {@arg_z@}\), since the real part is \(0\).
+Now we can write \(z\) in polar form: \[z = {@radius@} \, e^{ {@arg_z@} \, \mathrm i }\]
+ +b) Using the following formula, we can then calculate the complex \(n\):th roots:
+\[\sqrt[n]{z} = \sqrt[n]{r} \,\exp\left( \mathrm i\,\frac{\theta + 2k\pi}{n}\right), \quad k = 0, 1, \dots , n-1.\]
+ +With \(k=0\) we get +\[\begin{split} u_1 &= \sqrt[3]{{@radius@}} \exp\left( \mathrm i\, \frac{{@arg_z@}}{3} \right) \\ +&= {@polarform_simp(u1)@}. +\end{split}\]
+ +Similarly with \(k=1\) we get +\[\begin{split} u_2 &= \sqrt[3]{{@radius@}} \exp \left( \mathrm i\, \frac{{@arg_z@} + 2\pi}{3}\right) \\ +&= {@polarform_simp(u2)@}, +\end{split}\] +
+and with \(k=2\): +\[\begin{split} u_3 &= \sqrt[3]{{@radius@}}\exp \left( \mathrm i \, \frac{{@arg_z@} + 4\pi}{3}\right) \\ +&= {@polarform_simp(u3)@}. \end{split}\]
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-b)
+[[feedback:prt1]]
+b)
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]]>If \(z = {@zPar@}\), then what must the distance to the origin be? That distance is the same as the radius! (As usual, use the Pythagorean theorem to calculate the distance.)
]]>Use the figure! You know that the real part of \(z\) is \(0\), so what must the angle be? Remember the requirement \(-\pi < \theta \leq \pi\).
]]>null for all entries.
There do not exist \(2 \times 2 \) matrices \(E, \, F\) with rank\((E)\)\(=\)rank\((F)\)\(=2\) such that \(E\cdot F = \left( \begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array} \right) \).
-Select whether it is true or false, and give either a proof or counterexample as appropriate.
-Proof:
[[input:a4_true]][[validation:a4_true]] - [[/reveal]] - [[reveal input="a4" value="false"]] -Counterexample:
-There do not exist \(2 \times 2 \) matrices \(E, \, F\) with rank\((E)\)\(=\)rank\((F)\)\(=2\) such that \(E\cdot F = \left( \begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array} \right) \).
+Select whether it is true or false, and give either a proof or counterexample as appropriate.
+Proof:
[[input:a4_true]][[validation:a4_true]] + [[/reveal]] + [[reveal input="a4" value="false"]] +Counterexample:
+We denote the row of matrix \(A\) by \(A_i\), columns of matrix \(B\) by \(b_j\) and columns of matrix \(AB\) by \((ab)_j\). Then the matrix \(AB={@zmatrix@}\) are made up of two column vectors, namely \((ab)_1={@cvec@}\) and \((ab)_2={@cvec@}\).
-Thus, we break this problem down into two parts which are \((ab)_1=A \cdot b_1={@cvec@}\) and \((ab)_2=A \cdot b_2={@cvec@}\). -Note that this is a linear transformation from \(\mathbb{R}^2\) to \(\mathbb{R}^2\) as \(A\) maps \(b_1\) to \({@cvec@}\) and \(b_2\) to \({@cvec@}\). - -By the dimension theorem, \(\text{dim}(\mathbb{R}^2)= \text{nullity}(A)+\text{rank}(A)\). -As \(A\neq{@zmatrix@}\), \(\text{nullity}(A)<2\). On the other hand, \(b_1\neq{@cvec@}\) and \(b_2\neq{@cvec@}\) is in the null space of \(A\) so, \(\text{nullity}(A)\neq0\) which implies that \(\text{nullity}(A)=1\). This means that the rows or columns of the matrix \(A\) are linearly dependent. -
--Again, from the dimension theorem, we know that \(\text{rank}(B)\neq2\) since if \(\text{rank}(B)=2\) then \(\text{nullity}(A)=2\) which contradicts our assumption that \(A\) cannot be the zero matrix. From our assumption, we also know that \(\text{rank}(B)\neq0\) as \(B\) cannot be the zero matrix. Hence, similar to \(A\), \(B\) has \(\text{rank}=1\).
--To find an example of matrices \(A,B\) which satisfies the condition, we set \(A\) to be any matrix with linearly dependent columns or rows. Then solve for \(B\). -For example, \(A\) can be of the form \(\left( \begin{array}{cc} a & a \\ ca & ca \end{array} \right)\) for any \(c,a \in \mathbb{R}\).
+ We denote the row of matrix \(A\) by \(A_i\), columns of matrix \(B\) by \(b_j\) and columns of matrix \(AB\) by \((ab)_j\). Then the matrix \(AB={@zmatrix@}\) are made up of two column vectors, namely \((ab)_1={@cvec@}\) and \((ab)_2={@cvec@}\).
By the dimension theorem, \(\text{dim}(\mathbb{R}^2)= \text{nullity}(A)+\text{rank}(A)\). +As \(A\neq{@zmatrix@}\), \(\text{nullity}(A)<2\). On the other hand, \(b_1\neq{@cvec@}\) and \(b_2\neq{@cvec@}\) is in the null space of \(A\) so, \(\text{nullity}(A)\neq0\) which implies that \(\text{nullity}(A)=1\). This means that the rows or columns of the matrix \(A\) are linearly dependent. +
++Again, from the dimension theorem, we know that \(\text{rank}(B)\neq2\) since if \(\text{rank}(B)=2\) then \(\text{nullity}(A)=2\) which contradicts our assumption that \(A\) cannot be the zero matrix. From our assumption, we also know that \(\text{rank}(B)\neq0\) as \(B\) cannot be the zero matrix. Hence, similar to \(A\), \(B\) has \(\text{rank}=1\).
++To find an example of matrices \(A,B\) which satisfies the condition, we set \(A\) to be any matrix with linearly dependent columns or rows. Then solve for \(B\). +For example, \(A\) can be of the form \(\left( \begin{array}{cc} a & a \\ ca & ca \end{array} \right)\) for any \(c,a \in \mathbb{R}\).
One such example is \(A={@ta1A@}\). An example for \(B\) which satisfies the condition would be \(B={@ta1B@}\).
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Please give your answer to \(4\) significant figures.
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