Convert between any forms of complex numbers - High school Algebra II Exercises

Convert z=3+2i into the exponential form.

z=4e^{\frac{i\pi}{7}}

z=2.e^{i\frac{\pi}3}

z=3e^{\frac{i\pi}{6}}

z=\sqrt{13}e^{i tan^{-1}\left(\frac{2}{3}\right)}

If z=x+iy with x\neq0, then:

|z|=\sqrt{x^2+y^2}
\theta=tan^{-1}\left(\dfrac{y}{x}\right)

The exponential form of z is:

z=|z|e^{i\theta}

Here, we have:

z_1 = 3 + 2i

Therefore:

|z_1|=\sqrt{2^2+3^2}=\sqrt{13}
\theta=tan^{-1}\left(\dfrac{2}{3}\right)

The exponential form of z is:

z = \sqrt {13} e^{i \tan^{- 1} \left(\frac {2} {3}\right)}

Convert z=-1+2i into the exponential form.

\sqrt {6} e^{i \tan^{- 1} \left(2\right)}

\sqrt {11} e^{i \tan^{- 1} \left(-.5\right)}

\sqrt {5} e^{i \tan^{- 1} \left(-2\right)}

\sqrt {7} e^{i \tan^{- 1} \left(.5\right)}

If z=x+iy with x\neq0, then:

|z|=\sqrt{x^2+y^2}
\theta=tan^{-1}\left(\dfrac{y}{x}\right)

The exponential form of z is:

z=|z|e^{i\theta}

Here, we have:

z_1 = -1 + 2i

Therefore:

|z_1|=\sqrt{\left(-1\right)^2+2^2}=\sqrt{5}
\theta=tan^{-1}\left(\dfrac{2}{-1}\right)

The exponential form of z is:

z = \sqrt {5} e^{i \tan^{- 1} \left(-2\right)}

Convert z=7+4i into the exponential form.

\sqrt {65} e^{i \tan^{- 1} \left(\frac {4} {7}\right)}

\sqrt {67} e^{i \tan^{- 1} \left(\frac {4} {9}\right)}

\sqrt {59} e^{i \tan^{- 1} \left(\frac {4} {9}\right)}

\sqrt {61} e^{i \tan^{- 1} \left(\frac {4} {9}\right)}

If z=x+iy with x\neq0, then:

|z|=\sqrt{x^2+y^2}
\theta=tan^{-1}\left(\dfrac{y}{x}\right)

The exponential form of z is:

z=|z|e^{i\theta}

Here, we have:

z_1 = 7 + 4i

Therefore:

|z_1|=\sqrt{7^2+4^2}=\sqrt{65}
\theta=tan^{-1}\left(\dfrac{4}{7}\right)

The exponential form of z is:

z = \sqrt {65} e^{i \tan^{- 1} \left(\frac {4} {7}\right)}

Convert z=-4-5i into the exponential form.

\sqrt {47} e^{i \tan^{- 1} \left(\frac {7} {4}\right)}

\sqrt {43} e^{i \tan^{- 1} \left(\frac {9} {4}\right)}

\sqrt {41} e^{i \tan^{- 1} \left(\frac {5} {4}\right)}

\sqrt {29} e^{i \tan^{- 1} \left(\frac {-5} {4}\right)}

If z=x+iy with x\neq0, then:

|z|=\sqrt{x^2+y^2}
\theta=tan^{-1}\left(\dfrac{y}{x}\right)

The exponential form of z is:

z=|z|e^{i\theta}

Here, we have:

z_1 = -4 -5i

Therefore:

|z_1|=\sqrt{\left(-4\right)^2+\left(-5\right)^2}=\sqrt{41}
\theta=tan^{-1}\left(\dfrac{-5}{-4}\right)=tan^{-1}\left(\dfrac{5}{4}\right)

The exponential form of z is:

z = \sqrt {41} e^{i \tan^{- 1} \left(\frac {5} {4}\right)}

Convert z=-9+7i into the exponential form.

\sqrt {109} e^{i \tan^{- 1} \left(\frac {5} {9}\right)}

z = \sqrt {130} e^{i \tan^{- 1} \left(\frac {7} {11}\right)}

\sqrt {130} e^{i \tan^{- 1} \left(\frac {-7} {9}\right)}

\sqrt {119} e^{i \tan^{- 1} \left(\frac {7} {9}\right)}

If z=x+iy with x\neq0, then:

|z|=\sqrt{x^2+y^2}
\theta=tan^{-1}\left(\dfrac{y}{x}\right)

The exponential form of z is:

z=|z|e^{i\theta}

Here, we have:

z_1 =-9+7i

Therefore:

|z_1|=\sqrt{\left(-9\right)^2+7^2}=\sqrt{130}
\theta=tan^{-1}\left(\dfrac{7}{-9}\right)

The exponential form of z is:

z = \sqrt {130} e^{i \tan^{- 1} \left(\frac {-7} {9}\right)}

Convert z=7-7i into the exponential form.

7\sqrt {2} e^{\frac{-i \pi}{3} }

7\sqrt {3} e^{\frac{-i \pi}{5} }

7\sqrt {2} e^{\frac{-i \pi}{4} }

7\sqrt {3} e^{\frac{-i \pi}{6} }

If z=x+iy with x\neq0, then:

|z|=\sqrt{x^2+y^2}
\theta=tan^{-1}\left(\dfrac{y}{x}\right)

The exponential form of z is:

z=|z|e^{i\theta}

Here, we have:

z_1 = 7-7i

Therefore:

|z_1|=\sqrt{7^2+\left(-7\right)^2}=\sqrt{98}=7\sqrt{2}
\theta=tan^{-1}\left(\dfrac{-7}{7}\right)=tan^{-1}\left(-1\right)=-\dfrac{\pi}{4}

The exponential form of z is:

z = 7\sqrt {2} e^{\frac{-i \pi}{4} }

Convert z=1+\sqrt{3}i into the exponential form.

2 e^{i \frac{\pi}{3}}

3 e^{i \frac{\pi}{4}}

3 e^{i \frac{\pi}{5}}

2 e^{i \frac{\pi}{4}}

If z=x+iy with x\neq0, then:

|z|=\sqrt{x^2+y^2}
\theta=tan^{-1}\left(\dfrac{y}{x}\right)

The exponential form of z is:

z=|z|e^{i\theta}

Here, we have:

z_1 =1+\sqrt{3}i

Therefore:

|z_1|=\sqrt{1^2+\left(\sqrt{3}\right)^2}=\sqrt{4}=2
\theta=tan^{-1}\left(\dfrac{\sqrt{3}}{1}\right)=\dfrac{\pi}{3}

The exponential form of z is:

z =2 e^{i \frac{\pi}{3}}