Imaginary numbers | Solutions with procedure

9 Nov 2019 Leave a Comment

Here are the solutions to the exercises I gave you here!

$2i+6i-\frac{9i\cdot 3}{i^2}$ $=2i+6i-\frac{27i}{-1}=2i+6i+27i=35i$

$8i+2(i-4)^2$ $=8i+2\left ( -1-8i+16 \right )=8i+2(15-8i)=8i+30-16i=30-8i$

$i(i+4i^4)^2-\sqrt{-8}$ $=i(i+4)^2-2i\sqrt2=i(-1+8i+16)-2i\sqrt2$ $=-i-8+16-2i\sqrt2=-i\left ( 1+2\sqrt2 \right )+8=8-i\left ( 1+2\sqrt2 \right )$

$i^4\sqrt{-9i^2}$ $=1\sqrt{-9(-1)}=\sqrt9=3$

$\varphi+\eta\, i\sqrt{-\pi}$ $=\varphi+\eta\, i\cdot i\sqrt\pi=\varphi-\eta\sqrt\pi$

$i+\cos(i\sqrt{-\pi^2})$ $=i+\cos(i\cdot i\pi)=i+\cos(-\pi)=i+\cos(\pi)=i-1$

Notice that cosine is an even function, therefore cos(-pi)=cos(pi).

$\log_3(2i^4+1)$ $=\log_3(2\cdot 1+1)=\log_3(3)=1$

The logarithm in base 3 of 3 is 1 because 1 is the number you must raise 3 to in order to get 3.

$\pi^{\varphi+i^2}$ $=\pi^{\varphi +(-1)}=\pi^{\varphi-1}=\pi^\varphi\pi^{-1}=\frac{\pi^\varphi}{\pi}$

$\sqrt{-12}\, \theta^\log_\theta(-i)$ $=i\sqrt{4\cdot 3}\cdot (-i)=2i\sqrt3\cdot (-i)=-(-2)\sqrt3=2\sqrt3$

Now, why is theta raised to the log in base theta of negative i equal to negative i, i.e. the argument of the logarithm? That’s a property:

$a^{\log_a(x)}=x$

Another property is the following:

$\log_a(a)=1$

$\left | 5i\sqrt{-50} \right |$ $\left |5i\cdot i\sqrt{25\cdot 2} \right |=\left | -5\cdot 5\sqrt2 \right |=\left | -25\sqrt2 \right |=25\sqrt2$

Hope this was useful! If you have any doubt leave a comment and I will be happy to help!

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Imaginary and complex numbers Calculus, imaginary numbers exercises, Solutions

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Imaginary numbers | Solutions with procedure

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Imaginary numbers | Solutions with procedure

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