Solve {left(frac{4+3texttt{i}}{3-4texttt{i}}right)}^10+left|1-3texttt{i}right|

$${ \left( \frac{ 4+3 \texttt{i} }{ 3-4 \texttt{i} } \right) }^{ 10 } + \left| 1-3 \texttt{i} \right|$$

Evaluate

$$\sqrt{10}-1\approx 2.16227766$$

Solution Steps

Multiply both numerator and denominator of \frac{4+3i}{3-4i} by the complex conjugate of the denominator, 3+4i.

$$\left(\frac{\left(4+3i\right)\left(3+4i\right)}{\left(3-4i\right)\left(3+4i\right)}\right)^{10}+|1-3i|$$

Do the multiplications in \frac{\left(4+3i\right)\left(3+4i\right)}{\left(3-4i\right)\left(3+4i\right)}.

$$\left(\frac{25i}{25}\right)^{10}+|1-3i|$$

Divide 25i by 25 to get i.

$$i^{10}+|1-3i|$$

Calculate i to the power of 10 and get -1.

$$-1+|1-3i|$$

The modulus of a complex number a+bi is \sqrt{a^{2}+b^{2}}. The modulus of 1-3i is \sqrt{10}.

$$-1+\sqrt{10}$$

Real Part

$$\sqrt{10} - 1 = 2.16227766$$