Answer:
See below for answers and explanations
Step-by-step explanation:
Problem 1
Let [tex]z=r(\cos\theta+i\sin\theta)[/tex] be a complex number, where [tex]n[/tex] is an integer and [tex]n\geq 1[/tex]. By DeMoivre's Theorem, if [tex]z^n=r^n(\cos\theta+i\sin\theta)^n[/tex], then [tex]z^n=r^n(\cos n\theta+i\sin n\theta)[/tex].
Hence, we must first convert from rectangular to polar form:
[tex]z=-1-i\\\\r=\sqrt{a^2+b^2}\\\\r=\sqrt{(-1)^2+(-1)^2}\\\\r=\sqrt{1+1}\\\\r=\sqrt{2}[/tex]
[tex]\displaystyle \theta=\tan^{-1}\biggr(\frac{y}{x}\biggr)\\\\\theta=\tan^{-1}\biggr(\frac{-1}{-1}\biggr)\\ \\\theta=\tan^{-1}(1)\\\\\theta=45^\circ[/tex]
Therefore, the rectangular form for the complex number is [tex]z=\sqrt{2}(\cos45^\circ+i\sin45^\circ)[/tex], and now we can apply DeMoivre's Theorem to raise the complex number to the 10th power:
[tex]z=\sqrt{2}(\cos45^\circ+i\sin45^\circ)\\\\z^{10}=\sqrt{2}^{10}(\cos45^\circ+i\sin45^\circ)^{10}\\\\z^{10}=2^5(\cos10\cdot45^\circ+i\sin10\cdot45^\circ)\\\\z^{10}=32(\cos450^\circ+i\sin450^\circ)\\\\z^{10}=32(0+i)\\\\z^{10}=32i[/tex]
Therefore, [tex]z^{10}=32i[/tex] in rectangular form.
Problem 2
[tex]3\:\text{cis}\:90^\circ=3(\cos90^\circ+i\sin90^\circ)=3(0+i)=3i[/tex]
Therefore, [tex]3\:\text{cis}\:90^\circ=3i[/tex] in rectangular form.
