Sum of Squares

Clearly, $a^2 + b^2$ is the magnitude of the the complex number $a + bi$ , which is the same as the magnitude (also known as the norm ) of the complex number on the right hand side. Let this complex number be $z = \frac {u}{v}$ .

It is a property of complex numbers that $\left| \frac {u}{v} \right| = \dfrac {\left| u \right|}{\left| v \right|}.$ The same applies for multiplication. Since the magnitude of $1 + 3i$ is $\sqrt {10}$ and the mangitude of $6 + 8i$ is $10$ , the answer is

$\dfrac {\left( \sqrt {10} \right)^{60}}{\left( 10 \right)^{30}} = \boxed {1},$ which is our answer. $\Box$

We have:

$a+bi=\frac { { (1+3i })^{ 60 } }{ { (6+8i })^{ 30 } }$ $eq.1$ $\Longleftrightarrow$

$\Longleftrightarrow$ $a-bi=\frac { { (1-3i })^{ 60 } }{ { (6-8i })^{ 30 } }$ $eq.2$

By multiplying eq.1 and eq.2 we get:

$a^2+b^2=\frac { { (1+3i })^{ 60 } }{ { (6+8i })^{ 30 } } \times \frac { { (1-3i })^{ 60 } }{ { (6-8i })^{ 30 } } =\frac { { [(1+3i)\cdot (1-3i)] }^{ 60 } }{ { [(6+8i)\cdot (6-8i)] }^{ 30 } } =\quad \frac { { ({ 1 }^{ 2 }+{ 3 }^{ 2 }) }^{ 60 } }{ { ({ 6 }^{ 2 }+8^{ 2 }) }^{ 30 } } =\frac { { 10 }^{ 60 } }{ { 100 }^{ 30 } } =\frac { { 10 }^{ 60 } }{ { 10 }^{ 60 } } =\boxed { 1 }$

$\large{\frac{(1+3i)^{60}}{(6+8i)^{30}}=(\frac{(1+3i)^2}{6+8i})^{30}=(\frac{-8+6i}{6+8i})^{30}=(\frac{-8}{6+8i}+\frac{6i}{6+8i})^{30}=}$

$\large{(\frac{-12+16i}{25}+\frac{12+9i}{25})^{30}=i^{30}=-1\Rightarrow (-1)^2=\boxed{1}}.$