Binomial formula on complex number

Question

"Write the complex number $$z=(1+i)^9$$ on rectangular form"

How is this easiest solved? I guess the binomial formula could but used here, but I can't really seem to work out how to apply it.

Answer 1

Hint:

$$1+i = \sqrt{2}\exp \left(i \frac{\pi}{4} \right)$$

Answer 2

HINT: $$(1+i)^2=1-1+2i=2i$$ and we have $$2^{18}i^{18}$$

Answer 3

To evaluate the power of a complex number usually it is better to use the exponential notation (like in Siong Thye Goh's answer). If the exponent is low, like in this case, you may try in this way: $$(1+i)^2=1+i+i+i^2=2i\Rightarrow (1+i)^4=(2i)^2=4i^2=-4.$$ What is $(1+i)^8$? What is $(1+i)^{9}=(1+i)^8\cdot (1+i)$?

Answer 4

Polar form is the way to go here, then you can convert back. $$ 1+i=\sqrt{2}e^{\frac{\pi}{4}i}\implies (1+i)^9=2^{9/2}e^{\frac{9\pi}{4}i}\\ =\sqrt{2}2^4e^{\frac{\pi}{4}i} $$ which we put back in rectangular form by solving the system $$ x^2+y^2=2^9\\ x/y=\tan(\pi/4)=1 $$ and so $y=x$ and $$ 2x^2=2^9\implies x=2^4=y $$ and your complex number is really $$ 16+16i $$

Answer 5

$(1+i)^9$
$=1+9i+36i^2+84i^3+126i^4+126i^5+84i^6+36i^7+9i^8+i^9$ $=1+9i-36-84i+126+126i-84-36i+9+i$
$=(1-36+126-84+9)+(9-84+126-36+1)i$
$=16+16i.$

$1+i=\sqrt2(\cos45^\circ+i\sin45^\circ);$

$(1+i)^9=(\sqrt2)^9(\cos405^\circ+i\sin405^\circ)=16\sqrt2(\cos45^\circ+i\sin45^\circ)$
$=16\sqrt2\left(\frac1{\sqrt2}+\frac1{\sqrt2}i\right)=16+16i.$