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Given the complex numbers [tex]\( z_1 = 15\left(\cos 90^\circ + i \sin 90^\circ\right) \)[/tex] and [tex]\( z_2 = 3\left(\cos 10^\circ + i \sin 10^\circ\right) \)[/tex], let's analyze the transformation to form [tex]\( \frac{z_1}{z_2} \)[/tex].
First, convert each complex number to its exponential form:
[tex]\[ z_1 = 15 \text{cis}\ 90^\circ \quad \text{and} \quad z_2 = 3 \text{cis}\ 10^\circ \][/tex]
where [tex]\(\text{cis}\ \theta = \cos \theta + i \sin \theta\)[/tex].
When we divide [tex]\( z_1 \)[/tex] by [tex]\( z_2 \)[/tex]:
[tex]\[ \frac{z_1}{z_2} = \frac{15 \text{cis}\ 90^\circ}{3 \text{cis}\ 10^\circ} = \left( \frac{15}{3} \right) \frac{\text{cis}\ 90^\circ}{\text{cis}\ 10^\circ} \][/tex]
The magnitude (modulus) of the quotient is:
[tex]\[ \left| \frac{z_1}{z_2} \right| = \frac{|z_1|}{|z_2|} = \frac{15}{3} = 5 \][/tex]
The argument (angle) of the quotient is found by subtracting the angles:
[tex]\[ \arg \left( \frac{z_1}{z_2} \right) = 90^\circ - 10^\circ = 80^\circ \][/tex]
Therefore, the quotient in polar form is:
[tex]\[ \frac{z_1}{z_2} = 5 \text{cis}\ 80^\circ = 5 \left( \cos 80^\circ + i \sin 80^\circ \right) \][/tex]
This gives us the quotient as [tex]\( 5 \left( \cos 80^\circ + i \sin 80^\circ \right) \)[/tex].
Now, with respect to the transformation:
- The magnitude of [tex]\( z_1 \)[/tex] is scaled by a factor of [tex]\( 5 \)[/tex] (since [tex]\( \frac{15}{3} = 5 \)[/tex]).
- The angle is rotated [tex]\( 80^\circ \)[/tex] degrees clockwise (since we need to subtract [tex]\( 10^\circ \)[/tex] from [tex]\( 90^\circ \)[/tex]).
So, summarizing the information:
The quotient of [tex]\( \frac{z_1}{z_2} \)[/tex] is [tex]\( 5 \left( \cos \left(80^\circ\right) + i \sin \left(80^\circ\right) \right) \)[/tex].
[tex]\( z_1 \)[/tex] is scaled by a factor of 5 and then rotated [tex]\( 80 \)[/tex] degrees clockwise.
First, convert each complex number to its exponential form:
[tex]\[ z_1 = 15 \text{cis}\ 90^\circ \quad \text{and} \quad z_2 = 3 \text{cis}\ 10^\circ \][/tex]
where [tex]\(\text{cis}\ \theta = \cos \theta + i \sin \theta\)[/tex].
When we divide [tex]\( z_1 \)[/tex] by [tex]\( z_2 \)[/tex]:
[tex]\[ \frac{z_1}{z_2} = \frac{15 \text{cis}\ 90^\circ}{3 \text{cis}\ 10^\circ} = \left( \frac{15}{3} \right) \frac{\text{cis}\ 90^\circ}{\text{cis}\ 10^\circ} \][/tex]
The magnitude (modulus) of the quotient is:
[tex]\[ \left| \frac{z_1}{z_2} \right| = \frac{|z_1|}{|z_2|} = \frac{15}{3} = 5 \][/tex]
The argument (angle) of the quotient is found by subtracting the angles:
[tex]\[ \arg \left( \frac{z_1}{z_2} \right) = 90^\circ - 10^\circ = 80^\circ \][/tex]
Therefore, the quotient in polar form is:
[tex]\[ \frac{z_1}{z_2} = 5 \text{cis}\ 80^\circ = 5 \left( \cos 80^\circ + i \sin 80^\circ \right) \][/tex]
This gives us the quotient as [tex]\( 5 \left( \cos 80^\circ + i \sin 80^\circ \right) \)[/tex].
Now, with respect to the transformation:
- The magnitude of [tex]\( z_1 \)[/tex] is scaled by a factor of [tex]\( 5 \)[/tex] (since [tex]\( \frac{15}{3} = 5 \)[/tex]).
- The angle is rotated [tex]\( 80^\circ \)[/tex] degrees clockwise (since we need to subtract [tex]\( 10^\circ \)[/tex] from [tex]\( 90^\circ \)[/tex]).
So, summarizing the information:
The quotient of [tex]\( \frac{z_1}{z_2} \)[/tex] is [tex]\( 5 \left( \cos \left(80^\circ\right) + i \sin \left(80^\circ\right) \right) \)[/tex].
[tex]\( z_1 \)[/tex] is scaled by a factor of 5 and then rotated [tex]\( 80 \)[/tex] degrees clockwise.
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